Physics Notes Free Essays

string(244) " lie in the equatorial plane of the earth because it must accelerate in a plane where the centre of Earth lies since the net orce exerted on the satellite is the Earth’s gravitational force, which is directed towards the centre of Earth\." Gravitation Gravitational field strength at a point is defined as the gravitational force per unit mass at that point. Newton’s law of gravitation: The (mutual) gravitational force F between two point masses M and m separated by a distance r is given by F =| GMm| (where G: Universal gravitational constant)| | r2| | or, the gravitational force of between two point masses is proportional to the product of their masses ; inversely proportional to the square of their separation. Gravitational field strength at a point is the gravitational force per unit mass at that point. We will write a custom essay sample on Physics Notes or any similar topic only for you Order Now It is a vector and its S. I. unit is N kg-1. By definition, g = F / m By Newton Law of Gravitation, F = GMm / r2 Combining, magnitude of g = GM / r2 Therefore g = GM / r2, M = Mass of object â€Å"creating† the field Example 1: Assuming that the Earth is a uniform sphere of radius 6. 4 x 106 m and mass 6. 0 x 1024 kg, find the gravitational field strength g at a point: (a) on the surface, g = GM / r2 = (6. 67 ? 10-11)(6. 0 x 1024) / (6. 4 x 106)2 = 9. 77ms-2 (b) at height 0. 50 times the radius of above the Earth’s surface. g = GM / r2 = (6. 67 ? 10-11)(6. 0 x 1024) / ( (1. 5 ? 6. 4 x 106)2 = 4. 34ms-2 Example 2: The acceleration due to gravity at the Earth’s surface is 9. 0ms-2. Calculate the acceleration due to gravity on a planet which has the same density but twice the radius of Earth. g = GM / r2 gP / gE = MPrE2 / MErP2 = (4/3) ? rP3rE2? P / (4/3) ? rE3rP2? E = rP / rE = 2 Hence gP = 2 x 9. 81 = 19. 6ms-2 Assuming that Earth is a uniform sphere of mass M. The magnitude of the gravitational force from E arth on a particle of mass m, located outside Earth a distance r from the centre of the Earth is F = GMm / r2. When a particle is released, it will fall towards the centre of the Earth, as a result of the gravitational force with an acceleration ag. FG = mag ag = GM / r2 Hence ag = g Thus gravitational field strength g is also numerically equal to the acceleration of free fall. Example 1: A ship is at rest on the Earth’s equator. Assuming the earth to be a perfect sphere of radius R and the acceleration due to gravity at the poles is go, express its apparent weight, N, of a body of mass m in terms of m, go, R and T (the period of the earth’s rotation about its axis, which is one day). At the North Pole, the gravitational attraction is F = GMEm / R2 = mgo At the equator, Normal Reaction Force on ship by Earth = Gravitational attraction – centripetal force N = mgo – mR? = mgo – mR (2? / T)2 Gravitational potential at a point is defined as the work done (by an external agent) in bringing a unit mass from infinity to that point (without changing its kinetic energy). ? = W / m = -GM / r Why gravitational potential values are always negative? As the gravitational force on the mass is attractive, the work done by an ext ag ent in bringing unit mass from infinity to any point in the field will be negative work {as the force exerted by the ext agent is opposite in direction to the displacement to ensure that ? KE = 0} Hence by the definition of negative work, all values of ? re negative. g = -| d? | = – gradient of ? -r graph {Analogy: E = -dV/dx}| | dr| | Gravitational potential energy U of a mass m at a point in the gravitational field of another mass M, is the work done in bringing that mass m {NOT: unit mass, or a mass} from infinity to that point. ; U = m ? = -GMm / r Change in GPE, ? U = mgh only if g is constant over the distance h; {; h;; radius of planet} otherwise, must use: ? U = m? f-m? i | Aspects| Electric Field| Gravitational Field| 1. | Quantity interacting with or producing the field| Charge Q| Mass M| 2. Definition of Field Strength| Force per unit positive charge E = F / q| Force per unit mass g = F / M| 3. | Force between two Point Charges or Masses| Coulomb’s Law: Fe = Q1Q2 / 4 or2| Newton’s Law of Gravitation: Fg = G (GMm / r2)| 4. | Field Strength of isolated Point Charge or Mass| E = Q / 4 or2| g = G (GM / r2)| 5. | Definition of Potential| Work done in bringing a unit positive charge from infinity to the point; V = W /Q| Work done in bringing a unit mass from infinity to the point; ? = W / M| 6. | Potential of isolated Point Charge or Mass| V = Q / 4 or| ? -G (M / r)| 7. | Change in Potential Energy| ? U = q ? V| ? U = m | Total Energy of a Satellite = GPE + KE = (-GMm / r) + ? (GMm / r) Escape Speed of a Satellite By Conservation of Energy, Initial KE| +| Initial GPE| =| Final KE| +| Final GPE| (? mvE2)| +| (-GMm / r)| =| (0)| +| (0)| Thus escape speed, vE = v(2GM / R) Note : Escape speed of an object is independent of its mass For a satellite in circular orbit, â€Å"the centripetal force is provided by the gravitational force† {Must always state what force is providing the centripetal force before following eqn is used! Henc e GMm / r2 = mv2 / r = mr? 2 = mr (2? / T)2 A satellite does not move in the direction of the gravitational force {ie it stays in its circular orbit} because: the gravitational force exerted by the Earth on the satellite is just sufficient to cause the centripetal acceleration but not enough to also pull it down towards the Earth. {This explains also why the Moon does not fall towards the Earth} Geostationary satellite is one which is always above a certain point on the Earth (as the Earth rotates about its axis. For a geostationary orbit: T = 24 hrs, orbital radius (; height) are fixed values from the centre of the Earth, ang velocity w is also a fixed value; rotates fr west to east. However, the mass of the satellite is NOT a particular value ; hence the ke, gpe, ; the centripetal force are also not fixed values {ie their values depend on the mass of the geostationary satellite. } A geostationary orbit must lie in the equatorial plane of the earth because it must accelerate in a p lane where the centre of Earth lies since the net orce exerted on the satellite is the Earth’s gravitational force, which is directed towards the centre of Earth. You read "Physics Notes" in category "Papers" {Alternatively, may explain by showing why it’s impossible for a satellite in a non-equatorial plane to be geostationary. } Thermal Physics Internal Energy: is the sum of the kinetic energy of the molecules due to its random motion ; the potential energy of the molecules due to the intermolecular forces. Internal energy is determined by the values of the current state and is independent of how the state is arrived at. You can read also Thin Film Solar Cell Thus if a system undergoes a series of changes from one state A to another state B, its change in internal energy is the same, regardless of which path {the changes in the p ; V} it has taken to get from A to B. Since Kinetic Energy proportional to temp, and internal energy of the system = sum of its Kinetic Energy and Potential Energy, a rise in temperature will cause a rise in Kinetic Energy and thus an increase in internal energy. If two bodies are in thermal equilibrium, there is no net flow of heat energy between them and they have the same temperature. NB: this does not imply they must have the same internal energy as internal energy depends also on the number of molecules in the 2 bodies, which is unknown here} Thermodynamic (Kelvin) scale of temperature: theoretical scale that is independent of the properties of any particular substance. An absolute scale of temp is a temp scale which does not depend on the property of any particular substance (ie the thermodynamic scale) Abs olute zero: Temperature at which all substances have a minimum internal energy {NOT: zero internal energy. } T/K = T/ °C + 273. 15, by definition of the Celsius scale. Specific heat capacity is defined as the amount of heat energy needed to produce unit temperature change {NOT: by 1 K} for unit mass {NOT: 1 kg} of a substance, without causing a change in state. c = Q / m? T Specific latent heat of vaporisation is defined as the amount of heat energy needed to change unit mass of a substance from liquid phase to gaseous phase without a change of temperature. Specific latent heat of fusion is defined as the amount of heat energy needed to change unit mass of a substance from solid phase to liquid phase without a change of temperature L = Q / m {for both cases of vaporisation ; melting} The specific latent heat of vaporisation is greater than the specific latent heat of fusion for a given substance because * During vaporisation, there is a greater increase in volume than in fusion, * Thus more work is done against atmospheric pressure during vaporisation, * The increase in vol also means the INCREASE IN THE (MOLECULAR) POTENTIAL ENERGY, ; hence, internal energy, during vaporisation more than that during melting, * Hence by 1st Law of Thermodynamics, heat supplied during vaporisation more than that during melting; hence lv ; lf {since Q = ml = ? U – W}. Note: 1. the use of comparative terms: greater, more, and; 2. the increase in internal energy is due to an increase in the PE, NOT KE of molecules 3. the system here is NOT to be considered as an ideal gas system Similarly, you need to explain why, when a liq is boiling, thermal energy is being supplied, and yet, the temp of the liq does not change. | Melting| Boiling| Evaporation| Occurrence| Throughout the substance, at fixed temperature and pressure| On the surface, at all temperatures| Spacing(vol) ; PE of molecules| Increase slightly| Increase significantly| | Temperature ; hence KE of molecules| Remains constant during process| Decrease for remaining liquid| First Law of Thermodynamics: The increase in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system. ?U = W + Q| ? U: Increase in internal energy of the system Q: Heat supplied to the system W: work done on the system| {Need to recall the sign convention for all 3 terms} Work is done by a gas when it expands; work is done on a gas when it is ompressed. W = area under pressure – volume graph. For constant pressure {isobaric process}, Work done = pressure x ? Volume Isothermal process: a process where T = const {? U = 0 for ideal gas} ? U for a cycle = 0 {since U ? T, ; ? T = 0 for a cycle } Equation of state for an ideal gas: p V = n R T, where T is in Kelvin {NOT:  °C}, n: no. of moles. p V = N k T, where N: no. of molecules, k:Boltzmann con st Ideal Gas: a gas which obeys the ideal gas equation pV = nRT FOR ALL VALUES OF P, V ; T Avogadro constant: defined as the number of atoms in 12g of carbon-12. It is thus the number of particles (atoms or molecules) in one mole of substance. For an ideal gas, internal energy U = Sum of the KE of the molecules only {since PE = 0 for ideal gas} U = N x? m ;c2; = N x (3/2)kT {for monatomic gas} * U depends on T and number of molecules N * U ? T for a given number of molecules Ave KE of a molecule, ? m ;c2; ? T {T in K: not  °C} Dynamics Newton’s laws of motion: Newton’s First Law Every body continues in a state of rest or uniform motion in a straight line unless a net (external) force acts on it. Newton’s Second Law The rate of change of momentum of a body is directly proportional to the net force acting on the body, and the momentum change takes place in the direction of the net force. Newton’s Third Law When object X exerts a force on object Y, object Y exerts a force of the same type that is equal in magnitude and opposite in direction on object X. The two forces ALWAYS act on different objects and they form an action-reaction pair. Linear momentum and its conservation: Mass: is a measure of the amount of matter in a body, ; is the property of a body which resists change in motion. Weight: is the force of gravitational attraction (exerted by the Earth) on a body. Linear momentum: of a body is defined as the product of its mass and velocity ie p = m v Impulse of a force (I): is defined as the product of the force and the time ? t during which it acts ie I = F x ? t {for force which is const over the duration ? t} For a variable force, the impulse I = Area under the F-t graph { ? Fdt; may need to â€Å"count squares†} Impulse is equal in magnitude to the change in momentum of the body acted on by the force. Hence the change in momentum of the body is equal in mag to the area under a (net) force-time graph. {Incorrect to define impulse as change in momentum} Force: is defined as the rate of change of momentum, ie F = [ m (v – u) ] / t = ma or F = v dm / dt The {one} Newton: is defined as the force needed to accelerate a mass of 1 kg by 1 m s-2. Principle of Conservation of Linear Momentum: When objects of a system interact, their total momentum before and after interaction are equal if no net (external) force acts on the system. * The total momentum of an isolated system is constant m1 u1 + m2 u2 = m1 v1 + m2 v2 if net F = 0 {for all collisions } NB: Total momentum DURING the interaction/collision is also conserved. (Perfectly) elastic collision: Both momentum ; kinetic energy of the system are conserved. Inelastic collision: Only momentum is conserved, total kinetic energy is not conserved. Perfectly inelastic collision: Only momentum is conserved, and the particles stick togethe r after collision. (i. e. move with the same velocity. ) For all elastic collisions, u1 – u2 = v2 – v1 ie. relative speed of approach = relative speed of separation or, ? m1u12 + ? m2u22 = ? m1v12 + ? 2v22 In inelastic collisions, total energy is conserved but Kinetic Energy may be converted into other forms of energy such as sound and heat energy. Current of Electricity Electric current is the rate of flow of charge. {NOT: charged particles} Electric charge Q passing a point is defined as the product of the (steady) current at that point and the time for which the current flows, Q = I t One coulomb is defined as the charge flowing per second pass a point at which the current is one ampere. Example 1: An ion beam of singly-charged Na+ and K+ ions is passing through vacuum. If the beam current is 20 ? A, calculate the total number of ions passing any fixed point in the beam per second. (The charge on each ion is 1. 6 x 10-19 C. ) Current, I = Q / t = Ne / t where N is the no. of ions and e is the charge on one ion. No. of ions per second = N / t = I / e = (20 x 10-6) / (1. 6 x 10-19) = 1. 25 x 10-14 Potential difference is defined as the energy transferred from electrical energy to other forms of energy when unit charge passes through an electrical device, V = W / Q P. D. = Energy Transferred / Charge = Power / Current or, is the ratio of the power supplied to the device to the current flowing, V = P / I The volt: is defined as the potential difference between 2 pts in a circuit in which one joule of energy is converted from electrical to non-electrical energy when one coulomb passes from 1 pt to the other, ie 1 volt = One joule per coulomb Difference between Potential and Potential Difference (PD): The potential at a point of the circuit is due to the amount of charge present along with the energy of the charges. Thus, the potential along circuit drops from the positive terminal to negative terminal, and potential differs from points to points. Potential Difference refers to the difference in potential between any given two points. For example, if the potential of point A is 1 V and the potential at point B is 5 V, the PD across AB, or VAB , is 4 V. In addition, when there is no energy loss between two points of the circuit, the potential of these points is same and thus the PD across is 0 V. Example 2: A current of 5 mA passes through a bulb for 1 minute. The potential difference across the bulb is 4 V. Calculate: (a) The amount of charge passing through the bulb in 1 minute. Charge Q = I t = 5 x 10-3 x 60 = 0. 3 C (b) The work done to operate the bulb for 1 minute. Potential difference across the bulb = W / Q 4 = W / 0. Work done to operate the bulb for 1 minute = 0. 3 x 4 = 1. 2 J Electrical Power, P = V I = I2 / R = V2 / R {Brightness of a lamp is determined by the power dissipated, NOT: by V, or I or R alone} Example 3: A high-voltage transmission line with a resistance of 0. 4 ? km-1 carries a current of 500 A. The line is at a potential of 1200 kV at the power station and carries the current to a city lo cated 160 km from the power station. Calculate (a) the power loss in the line. The power loss in the line P = I2 R = 5002 x 0. 4 x 160 = 16 MW (b) the fraction of the transmitted power that is lost. The total power transmitted = I V = 500 x 1200 x 103 = 600 MW The fraction of power loss = 16 / 600 = 0. 267 Resistance is defined as the ratio of the potential difference across a component to the current flowing through it , R = VI {It is NOT defined as the gradient of a V-I graph; however for an ohmic conductor, its resistance equals the gradient of its V-I graph as this graph is a straight line which passes through the origin} The Ohm: is the resistance of a resistor if there is a current of 1 A flowing through it when the pd across it is 1 V, ie, 1 ? = One volt per ampere Example 4: In the circuit below, the voltmeter reading is 8. 00 V and the ammeter reading is 2. 00 A. Calculate the resistance of R. Resistance of R = V / I = 8 / 2 = 4. 0 ? | | Temperature characteristics of thermistors: The resistance (i. e. the ratio V / I) is constant because metallic conductors at constant temperature obey Ohm’s Law. | As V increases, the temperature increases, resulting in an increase in the amplitude of vibration of ions and the collision frequency of electrons with the lattice ions. Hence the resistance of the filament increases with V. | A thermistor is made from semi-conductors. As V increases, temperature increases. This releases more charge carriers (electrons and holes) from the lattice, thus reducing the resistance of the thermistor. Hence, resistance decreases as temperature increases. | In forward bias, a diode has low resistance. In reverse bias, the diode has high resistance until the breakdown voltage is reached. | Ohm’s law: The current in a component is proportional to the potential difference across it provided physical conditions (eg temp) stay constant. R = ? L / A {for a conductor of length l, uniform x-sect area A and resistivity ? Resistivity is defined as the resistance of a material of unit cross-sectional area and unit length. {From R = ? l / A , ? = RA / L} Example 5: Calculate the resistance of a nichrome wire of length 500 mm and diameter 1. 0 mm, given that the resistivity of nichrome is 1. 1 x 10-6 ? m. Resistance, R = ? l / A = [(1. 1 x 10-6)(500 x 10-3)] / ? (1 x 10-3 / 2)2 = 0. 70 ? Electromotive force (Emf) is defined as t he energy transferred / converted from non-electrical forms of energy into electrical energy when unit charge is moved round a complete circuit. ie EMF = Energy Transferred per unit charge E = WQ EMF refers to the electrical energy generated from non-electrical energy forms, whereas PD refers to electrical energy being changed into non-electrical energy. For example, EMF Sources| Energy Change| PD across| Energy Change| Chemical Cell| Chem ; Elec| Bulb| Elec ; Light| Generator| Mech ; Elec| Fan| Elec ; Mech| Thermocouple| Thermal ; Elec| Door Bell| Elec ; Sound| Solar Cell| Solar ; Elec| Heating element| Elec ; Thermal| Effects of the internal resistance of a source of EMF: Internal resistance is the resistance to current flow within the power source. It reduces the potential difference (not EMF) across the terminal of the power supply when it is delivering a current. Consider the circuit below: The voltage across the resistor, V = IR, The voltage lost to internal resistance = Ir Thus, the EMF of the cell, E = IR + Ir = V + Ir Therefore If I = 0A or if r = 0? , V = E Motion in a Circle Kinematics of uniform circular motion Radian (rad) is the S. I. unit for angle, ? and it can be related to degrees in the following way. In one complete revolution, an object rotates through 360 ° , or 2? rad. As the object moves through an angle ? , with respect to the centre of rotation, this angle ? s known as the angular displacement. Angular velocity (? ) of the object is the rate of change of angular displacement with respect to time. ? = ? / t = 2? / T (for one complete revolution) Linear velocity, v, of an object is its instantaneous velocity at any point in its circular path. v = arc length / time taken = r? / t = r? * The direction of th e linear velocity is at a tangent to the circle described at that point. Hence it is sometimes referred to as the tangential velocity * ? is the same for every point in the rotating object, but the linear velocity v is greater for points further from the axis. A body moving in a circle at a constant speed changes velocity {since its direction changes}. Thus, it always experiences an acceleration, a force and a change in momentum. Centripetal acceleration a = r? 2 = v2 / r {in magnitude} Centripetal force Centripetal force is the resultant of all the forces that act on a system in circular motion. {It is not a particular force; â€Å"centripetal† means â€Å"centre-seeking†. Also, when asked to draw a diagram showing all the forces that act on a system in circular motion, it is wrong to include a force that is labelled as â€Å"centripetal force†. } Centripetal force, F = m r ? 2 = mv2 / r {in magnitude} A person in a satellite orbiting the Earth experiences â€Å"weightlessness† although the gravi field strength at that height is not zero because the person and the satellite would both have the same acceleration; hence the contact force between man ; satellite / normal reaction on the person is zero {Not because the field strength is negligible}. D. C. Circuits Circuit Symbols: Open Switch| Closed Switch| Lamp| Cell| Battery| Voltmeter| Resistor| Fuse| Ammeter| Variable resistor| Thermistor| Light dependent resistor (LDR)| Resistors in Series: R = R1 + R2 + †¦ Resistors in Parallel: 1/R = 1/R1 + 1/R2 + †¦ Example 1: Three resistors of resistance 2 ? , 3 ? and 4 ? respectively are used to make the combinations X, Y and Z shown in the diagrams. List the combinations in order of increasing resistance. Resistance for X = [1/2 + 1/(4+3)]-1 = 1. 56 ? Resistance for Y = 2 + (1/4 + 1/3)-1 = 3. 71 ? Resistance for Z = (1/3 + 1/2 + 1/4)-1 = 0. 923 ? Therefore, the combination of resistors in order of increasing resistance is Z X Y. Example: Referring to the circuit drawn, determine the value of I1, I and R, the combined resistance in the circuit. E = I1 (160) = I2 (4000) = I3 (32000) I1 = 2 / 160 = 0. 0125 A I2 = 2 / 4000 = 5 x 10-4 A I3 = 2 / 32000 = 6. 25 x 10-5 ASince I = I1 + I2 + I3, I = 13. 1 mAApplying Ohm’s Law, R = 213. 1 x 10-3 = 153 ? | | Example: A battery with an EMF of 20 V and an internal resistance of 2. 0 ? is connected to resistors R1 and R2 as shown in the diagram. A total current of 4. 0 A is supplied by the battery and R2 has a resistance of 12 ?. Calculate the resistance of R1 and the power supplied to each circuit component. E – I r = I2 R2 20 – 4 (2) = I2 (12) I2 = 1A Therefore, I1 = 4 – 1 = 3 AE – I r = I1 R1 12 = 3 R1 Therefore, R1 = 4Power supplied to R1 = (I1)2 R1 = 36 W Power supplied to R2 = (I2)2 R2 = 12 W| | For potential divider with 2 resistors in series, Potential drop across R1, V1 = R1 / (R1 + R2) x PD across R1 ; R2 Potential drop across R2, V1 = R2 / (R1 + R2) x PD across R1 ; R2 Example: Two resistors, of resistance 300 k? and 500 k? respectively, form a potential divider with outer junctions maintained at potentials of +3 V and -15 V. Determine the potential at the junction X between the resistors. The potential difference across the 300 k? resistor = 300 / (300 + 500) [3 – (-15)] = 6. 75 V The potential at X = 3 – 6. 75 = -3. 75 V A thermistor is a resistor whose resistance varies greatly with temperature. Its resistance decreases with increasing temperature. It can be used in potential divider circuits to monitor and control temperatures. Example: In the figure on the right, the thermistor has a resistance of 800 ? when hot, and a resistance of 5000 ? when cold. Determine the potential at W when the temperature is hot. When thermistor is hot, potential difference across it = [800 / (800 + 1700)] x (7 – 2) = 1. 6 VThe potential at W = 2 + 1. 6 V = 3. 6 V| | A Light dependent resistor (LDR) is a resistor whose resistance varies with the intensity of light falling on it. Its resistance decreases with increasing light intensity. It can be used in a potential divider circuit to monitor light intensity. Example: In the figure below, the resistance of the LDR is 6. 0 M in the dark but then drops to 2. 0 k in the light Determine the potential at point P when the LDR is in the light. In the light the potential difference across the LDR= [2k / (3k + 2k)] x (18 – 3) = 6 VThe potential at P = 18 – 6= 12 V| | The potential difference along the wire is proportional to the length of the wire. The sliding contact will move along wire AB until it finds a point along the wire such that the galvanometer shows a zero reading. When the galvanometer shows a zero reading, the current through the galvanometer (and the device that is being tested) is zero and the potentiometer is said to be â€Å"balanced†. If the cell has negligible internal resistance, and if the potentiometer is balanced, EMF / PD of the unknown source, V = [L1 / (L1 + L2)] x E Example: In the circuit shown, the potentiometer wire has a resistance of 60 ?. Determine the EMF of the unknown cell if the balanced point is at B. Resistance of wire AB= [0. 65 / (0. 65 + 0. 35)] x 60 = 39 ? EMF of the test cell= [39 / (60 + 20)] x 12| Work, Energy and Power Work Done by a force is defined as the product of the force and displacement (of its point of application) in the direction of the force W = F s cos ? Negative work is said to be done by F if x or its compo. is anti-parallel to F If a variable force F produces a displacement in the direction of F, the work done is determined from the area under F-x graph. {May need to find area by â€Å"counting the squares†. } By Principle of Conservation of Energy, Work Done on a system = KE gain + GPE gain + Work done against friction} Consider a rigid object of mass m that is initially at rest. To accelerate it uniformly to a speed v, a constant net force F is exerted on it, parallel to its motion over a displacement s. Since F is constant, acceleration is constant, Therefore, using the equation: v2 = u2 +2as, as = 12 (v2 – u2) Since kinetic energy is equal to the work done on the mass to bring it from rest to a speed v, The kinetic energy, EK| = Work done by the force F = Fs = mas = ? m (v2 – u2)| Gravitational potential energy: this arises in a system of masses where there are attractive gravitational forces between them. The gravitational potential energy of an object is the energy it possesses by virtue of its position in a gravitational field. Elastic potential energy: this arises in a system of atoms where there are either attractive or repulsive short-range inter-atomic forces between them. Electric potential energy: this arises in a system of charges where there are either attractive or repulsive electric forces between them. The potential energy, U, of a body in a force field {whether gravitational or electric field} is related to the force F it experiences by: F = – dU / dx. Consider an object of mass m being lifted vertically by a force F, without acceleration, from a certain height h1 to a height h2. Since the object moves up at a constant speed, F is equal to mg. The change in potential energy of the mass| = Work done by the force F = F s = F h = m g h| Efficiency: The ratio of (useful) output energy of a machine to the input energy. ie =| Useful Output Energy| x100% =| Useful Output Power| x100%| | Input Energy| | Input Power| | Power {instantaneous} is defined as the work done per unit time. P =| Total Work Done| =| W| | Total Time| | t| Since work done W = F x s, P =| F x s| =| Fv| | t| | | * for object moving at const speed: F = Total resistive force {equilibrium condition} * for object beginning to accelerate: F = Total resistive force + ma Forces Hooke’s Law: Within the limit of proportionality, the extension produced in a material is directly proportional to the force/load applied F = kx Force constant k = force per unit extension (F/x) Elastic potential energy/strain energy = Area under the F-x graph {May need to â€Å"count the squares†} For a material that obeys Hooke? s law, Elastic Potential Energy, E = ? F x = ? x2 Forces on Masses in Gravitational Fields: A region of space in which a mass experiences an (attractive) force due to the presence of another mass. Forces on Charge in Electric Fields: A region of space where a charge experiences an (attractive or repulsive) force due to the presence of another charge. Hydrostatic Pressure p = ? gh {or, pressure difference between 2 points separated by a vertical distance of h } Upthrust: An upward force exerted by a fluid on a submerged or floating object; arises because of the difference in pressure between the upper and lower surfaces of the object. Archimedes’ Principle: Upthrust = weight of the fluid displaced by submerged object. ie Upthrust = Volsubmerged x ? fluid x g Frictional Forces: * The contact force between two surfaces = (friction2 + normal reaction2)? * The component along the surface of the contact force is called friction * Friction between 2 surfaces always opposes relative motion {or attempted motion}, and * Its value varies up to a maximum value {called the static friction} Viscous Forces: * A force that opposes the motion of an object in a fluid * Only exists when there is (relative) motion Magnitude of viscous force increases with the speed of the object Centre of Gravity of an object is defined as that pt through which the entire weight of the object may be considered to act. A couple is a pair of forces which tends to produce rotation only. Moment of a Force: The product of the force and the perpendicular distance of its line of action to the pivot Torque of a Couple: The produce of one of the force s of the couple and the perpendicular distance between the lines of action of the forces. (WARNING: NOT an action-reaction pair as they act on the same body. ) Conditions for Equilibrium (of an extended object): 1. The resultant force acting on it in any direction equals zero 2. The resultant moment about any point is zero If a mass is acted upon by 3 forces only and remains in equilibrium, then 1. The lines of action of the 3 forces must pass through a common point 2. When a vector diagram of the three forces is drawn, the forces will form a closed triangle (vector triangle), with the 3 vectors pointing in the same orientation around the triangle. Principle of Moments: For a body to be in equilibrium, the sum of all the anticlockwise moments about any point must be equal to the sum of all the clockwise moments about that same point. Measurement Base quantities and their units; mass (kg), length (m), time (s), current (A), temperature (K), amount of substance (mol): Base Quantities| SI Units| | Name| Symbol| Length| metre| m| Mass| kilogram| kg| Time| second| s| Amount of substance| mole| mol| Temperature| Kelvin| K| Current| ampere| A| Luminous intensity| candela| cd| Derived units as products or quotients of the base units: Derived| Quantities Equation| Derived Units| Area (A)| A = L2| m2| Volume (V)| V = L3| m3| Density (? )| ? = m / V| kg m-3| Velocity (v)| v = L / t| ms-1| Acceleration (a)| a = ? v / t| ms-1 / s = ms-2| Momentum (p)| p = m x v| (kg)(ms-1) = kg m s-1| Derived Quantities| Equation| Derived Unit| Derived Units| | | Special Name| Symbol| | Force (F)| F = ? p / t| Newton| N| [(kg m s-1) / s = kg m s-2| Pressure (p)| p = F / A| Pascal| Pa| (kg m s-2) / m2 = kg m-1 s-2| Energy (E)| E = F x d| joule| J| (kg m s-2)(m) = kg m2 s-2| Power (P)| P = E / t| watt| W| (kg m2 s-2) / s = kg m2 s-3| Frequency (f)| f = 1 / t| hertz| Hz| 1 / s = s-1| Charge (Q)| Q = I x t| coulomb| C| A s| Potential Difference (V)| V = E / Q| volt| V| (kg m2 s-2) / A s = kg m2 s-3 A-1| Resistance (R)| R = V / I| ohm| ? (kg m2 s-3 A-1) / A = kg m2 s-3 A-2| Prefixes and their symbols to indicate decimal sub-multiples or multiples of both base and derived units: Multiplying Factor| Prefix| Symbol| 10-12| pico| p| 10-9| nano| n| 10-6| micro| ? | 10-3| milli| m| 10-2| centi| c| 10-1| decid| d| 103| kilo| k| 106| mega| M| 109| giga| G| 1012| tera| T| Estimates of physical quantities: When making an estimate, it is only reason able to give the figure to 1 or at most 2 significant figures since an estimate is not very precise. Physical Quantity| Reasonable Estimate| Mass of 3 cans (330 ml) of Coke| 1 kg| Mass of a medium-sized car| 1000 kg| Length of a football field| 100 m| Reaction time of a young man| 0. 2 s| * Occasionally, students are asked to estimate the area under a graph. The usual method of counting squares within the enclosed area is used. (eg. Topic 3 (Dynamics), N94P2Q1c) * Often, when making an estimate, a formula and a simple calculation may be involved. EXAMPLE 1: Estimate the average running speed of a typical 17-year-old? s 2. 4-km run. velocity = distance / time = 2400 / (12. 5 x 60) = 3. 2 ? 3 ms-1 EXAMPLE 2: Which estimate is realistic? | Option| Explanation| A| The kinetic energy of a bus travelling on an expressway is 30000J| A bus of mass m travelling on an expressway will travel between 50 to 80 kmh-1, which is 13. 8 to 22. 2 ms-1. Thus, its KE will be approximately ? m(182) = 162m. Thus, for its KE to be 30000J: 162m = 30000. Thus, m = 185kg, which is an absurd weight for a bus; ie. This is not a realistic estimate. | B| The power of a domestic light is 300W. | A single light bulb in the house usually runs at about 20W to 60W. Thus, a domestic light is unlikely to run at more than 200W; this estimate is rather high. | C| The temperature of a hot oven is 300 K. 300K = 27 0C. Not very hot. | D| The volume of air in a car tyre is 0. 03 m3. | | Estimating the width of a tyre, t, is 15 cm or 0. 15 m, and estimating R to be 40 cm and r to be 30 cm,volume of air in a car tyre is = ? (R2 – r2)t = ? (0. 42 – 0. 32)(0. 15) = 0. 033 m3 ? 0. 03 m3 (to one sig. fig. )| Distinction between systematic errors (including zero errors) an d random errors and between precision and accuracy: Random error: is the type of error which causes readings to scatter about the true value. Systematic error: is the type of error which causes readings to deviate in one direction from the true value. Precision: refers to the degree of agreement (scatter, spread) of repeated measurements of the same quantity. {NB: regardless of whether or not they are correct. } Accuracy: refers to the degree of agreement between the result of a measurement and the true value of the quantity. | ; ; R Error Higher ; ; ; ; ; ; Less Precise ; ; ;| v v vS Error HigherLess Accuratev v v| | | | | | Assess the uncertainty in a derived quantity by simple addition of actual, fractional or percentage uncertainties (a rigorous statistical treatment is not required). For a quantity x = (2. 0  ± 0. 1) mm, Actual/ Absolute uncertainty, ? x =  ± 0. 1 mm Fractional uncertainty, ? xx = 0. 05 Percentage uncertainty, ? xx 100% = 5 % If p = (2x + y) / 3 or p = (2x – y) / 3, ? p = (2? x + ? y) / 3 If r = 2xy3 or r = 2x / y3, ? r / r = ? x / x + 3? y / y Actual error must be recorded to only 1 significant figure, ; The number of decimal places a calculated quantity should have is determined by its actual error. For eg, suppose g has been initially calculated to be 9. 80645 ms-2 ; ? g has been initially calculated to be 0. 04848 ms-2. The final value of ? g must be recorded as 0. 5 ms-2 {1 sf }, and the appropriate recording of g is (9. 81  ± 0. 05) ms-2. Distinction between scalar and vector quantities: | Scalar| Vector| Definition| A scalar quantity has a magnitude only. It is completely described by a certain number and a unit. | A vector quantity has both magnitude and direction. It can be described by an arrow whose length represents the magnitude of the vector and the arrow-hea d represents the direction of the vector. | Examples| Distance, speed, mass, time, temperature, work done, kinetic energy, pressure, power, electric charge etc. Common Error: Students tend to associate kinetic energy and pressure with vectors because of the vector components involved. However, such considerations have no bearings on whether the quantity is a vector or scalar. | Displacement, velocity, moments (or torque), momentum, force, electric field etc. | Representation of vector as two perpendicular components: In the diagram below, XY represents a flat kite of weight 4. 0 N. At a certain instant, XY is inclined at 30 ° to the horizontal and the wind exerts a steady force of 6. 0 N at right angles to XY so that the kite flies freely. By accurate scale drawing| By calculations using sine and cosine rules, or Pythagoras? theorem| Draw a scale diagram to find the magnitude and direction of the resultant force acting on the kite. R = 3. 2 N (? 3. 2 cm) at ? = 112 ° to the 4 N vector. | Using cosine rule, a2 = b2 + c2 – 2bc cos A R2 = 42 + 62 -2(4)(6)(cos 30 °) R = 3. 23 NUsing sine rule: a / sin A = b / sin B 6 / sin ? = 3. 23 / sin 30 ° ? = 68 ° or 112 ° = 112 ° to the 4 N vector| Summing Vector Components| | Fx = – 6 sin 30 ° = – 3 NFy = 6 cos 30 ° – 4 = 1. 2 NR = v(-32 + 1. 22) = 3. 23 Ntan ? = 1. 2 / 3 = 22 °R is at an angle 112 ° to the 4 N vector. (90 ° + 22 °)| Kinematics Displacement, speed, velocity and acceleration: Distance: Total length covered irrespective of the direction of motion. Displacement: Distance moved in a certain direction. Speed: Distance travelled per unit time. Velocity: is defined as the rate of change of displacement, or, displacement per unit time {NOT: displacement over time, nor, displacement per second, nor, rate of change of displacement per unit time} Acceleration: is defined as the rate of change of velocity. Using graphs to find displacement, velocity and acceleration: * The area under a velocity-time graph is the change in displacement. The gradient of a displacement-time graph is the {instantaneous} velocity. * The gradient of a velocity-time graph is the acceleration. The ‘SUVAT’ Equations of Motion The most important word for this chapter is SUVAT, which stands for: * S (displacement), * U (initial velocity), * V (final velocity), * A (acceleration) and * T (time) of a particle that is in moti on. Below is a list of the equations you MUST memorise, even if they are in the formula book, memorise them anyway, to ensure you can implement them quickly. 1. v = u +at| derived from definition of acceleration: a = (v – u) / t| 2. | s = ? (u + v) t| derived from the area under the v-t graph| 3. | v2 = u2 + 2as| derived from equations (1) and (2)| 4. | s = ut + ? at2| derived from equations (1) and (2)| These equations apply only if the motion takes place along a straight line and the acceleration is constant; {hence, for eg. , air resistance must be negligible. } Motion of bodies falling in a uniform gravitational field with air resistance: Consider a body moving in a uniform gravitational field under 2 different conditions: Without Air Resistance: Assuming negligible air resistance, whether the body is moving up, or at the highest point or moving down, the weight of the body, W, is the only force acting on it, causing it to experience a constant acceleration. Thus, the gradient of the v-t graph is constant throughout its rise and fall. The body is said to undergo free fall. With Air Resistance: If air resistance is NOT negligible and if it is projected upwards with the same initial velocity, as the body moves upwards, both air resistance and weight act downwards. Thus its speed will decrease at a rate greater than . 81 ms-2 . This causes the time taken to reach its maximum height reached to be lower than in the case with no air resistance. The max height reached is also reduced. At the highest point, the body is momentarily at rest; air resistance becomes zero and hence the only force acting on it is the weight. The acceleration is thus 9. 81 ms-2 at this point. As a body falls, air resistance opposes its weight. The downward acceleration is thus less than 9. 81 ms-2. As air resistance increases with speed, it eventually equals its weight (but in opposite direction). From then there will be no resultant force acting on the body and it will fall with a constant speed, called the terminal velocity. Equations for the horizontal and vertical motion: | x direction (horizontal – axis)| y direction (vertical – axis)| s (displacement)| sx = ux t sx = ux t + ? ax t2| sy = uy t + ? ay t2 (Note: If projectile ends at same level as the start, then sy = 0)| u (initial velocity)| ux| uy| v (final velocity)| vx = ux + axt (Note: At max height, vx = 0)| vy = uy + at vy2 = uy2 + 2asy| a (acceleration)| ax (Note: Exists when a force in x direction present)| ay (Note: If object is falling, then ay = -g)| (time)| t| t| Parabolic Motion: tan ? = vy / vx ?: direction of tangential velocity {NOT: tan ? = sy / sx } Forces Hooke’s Law: Within the limit of proportionality, the extension produced in a material is directly proportional to the force/load applied F = kx Force constant k = force per unit extension (F/x) Elastic potential energy/strain ener gy = Area under the F-x graph {May need to â€Å"count the squares†} For a material that obeys Hooke? s law, Elastic Potential Energy, E = ? F x = ? k x2 Forces on Masses in Gravitational Fields: A region of space in which a mass experiences an (attractive) force due to the presence of another mass. Forces on Charge in Electric Fields: A region of space where a charge experiences an (attractive or repulsive) force due to the presence of another charge. Hydrostatic Pressure p = ? gh {or, pressure difference between 2 points separated by a vertical distance of h } Upthrust: An upward force exerted by a fluid on a submerged or floating object; arises because of the difference in pressure between the upper and lower surfaces of the object. Archimedes’ Principle: Upthrust = weight of the fluid displaced by submerged object. ie Upthrust = Volsubmerged x ? fluid x g Frictional Forces: The contact force between two surfaces = (friction2 + normal reaction2)? * The component along the surface of the contact force is called friction * Friction between 2 surfaces always opposes relative motion {or attempted motion}, and * Its value varies up to a maximum value {called the static friction} Viscous Forces: * A force that opposes the motion of an object in a fluid * Only exists when the re is (relative) motion * Magnitude of viscous force increases with the speed of the object Centre of Gravity of an object is defined as that pt through which the entire weight of the object may be considered to act. A couple is a pair of forces which tends to produce rotation only. Moment of a Force: The product of the force and the perpendicular distance of its line of action to the pivot Torque of a Couple: The produce of one of the forces of the couple and the perpendicular distance between the lines of action of the forces. (WARNING: NOT an action-reaction pair as they act on the same body. ) Conditions for Equilibrium (of an extended object): 1. The resultant force acting on it in any direction equals zero 2. The resultant moment about any point is zero If a mass is acted upon by 3 forces only and remains in equilibrium, then 1. The lines of action of the 3 forces must pass through a common point 2. When a vector diagram of the three forces is drawn, the forces will form a closed triangle (vector triangle), with the 3 vectors pointing in the same orientation around the triangle. Principle of Moments: For a body to be in equilibrium, the sum of all the anticlockwise moments about any point must be equal to the sum of all the clockwise moments about that same point. Work, Energy and Power Work Done by a force is defined as the product of the force and displacement (of its point of application) in the direction of the force W = F s cos ? Negative work is said to be done by F if x or its compo. is anti-parallel to F If a variable force F produces a displacement in the direction of F, the work done is determined from the area under F-x graph. {May need to find area by â€Å"counting the squares†. } By Principle of Conservation of Energy, Work Done on a system = KE gain + GPE gain + Work done against friction} Consider a rigid object of mass m that is initially at rest. To accelerate it uniformly to a speed v, a constant net force F is exerted on it, parallel to its motion over a displacement s. Since F is constant, acceleration is constant, Therefore, using the equation: 2 = u2 +2as, as = 12 (v2 – u2) Since kinetic energy is equal to the work done on the mass to bring it from rest to a speed v, The kinetic energy, EK| = Work done by the force F = Fs = mas = ? m (v2 – u2)| Gravitational potential energy: this arises in a system of masses where there are attractive gravitational forces between them. The gravitational potential energy of an object is the energy it possesses by virtue of its position in a gravitational field. Elastic potential energy: this arises in a system of atoms where there are either attractive or repulsive short-range inter-atomic forces between them. Electric potential energy: this arises in a system of charges where there are either attractive or repulsive electric forces between them. The potential energy, U, of a body in a force field {whether gravitational or electric field} is related to the force F it experiences by: F = – dU / dx. Consider an object of mass m being lifted vertically by a force F, without acceleration, from a certain height h1 to a height h2. Since the object moves up at a constant speed, F is equal to mg. The change in potential energy of the mass| = Work done by the force F = F s = F h = m g h| Efficiency: The ratio of (useful) output energy of a machine to the input energy. ie =| Useful Output Energy| x100% =| Useful Output Power| x100%| | Input Energy| | Input Power| | Power {instantaneous} is defined as the work done per unit time. P =| Total Work Done| =| W| | Total Time| | t| Since work done W = F x s, P =| F x s| =| Fv| | t| | | * for object moving at const speed: F = Total resistive force {equilibrium condition} * for object beginning to accelerate: F = Total resistive force + ma Wave Motion Displacement (y): Position of an oscillating particle from its equilibrium position. Amplitude (y0 or A): The maximum magnitude of the displacement of an oscillating particle from its equilibrium position. Period (T): Time taken for a particle to undergo one complete cycle of oscillation. Frequency (f): Number of oscillations performed by a particle per unit time. Wavelength (? ): For a progressive wave, it is the distance between any two successive particles that are in phase, e. g. it is the distance between 2 consecutive crests or 2 troughs. Wave speed (v): The speed at which the waveform travels in the direction of the propagation of the wave. Wave front: A line or surface joining points which are at the same state of oscillation, i. e. in phase, e. g. a line joining crest to crest in a wave. Ray: The path taken by the wave. This is used to indicate the direction of wave propagation. Rays are always at right angles to the wave fronts (i. e. wave fronts are always perpendicular to the direction of propagation). From the definition of speed, Speed = Distance / Time A wave travels a distance of one wavelength, ? , in a time interval of one period, T. The frequency, f, of a wave is equal to 1 / T Therefore, speed, v = ? / T = (1 / T)? f? v = f? Example 1: A wave travelling in the positive x direction is showed in the figure. Find the amplitude, wavelength, period, and speed of the wave if it has a frequency of 8. 0 Hz. Amplitude (A) = 0. 15 mWavelength (? ) = 0. 40 mPeriod (T) = 1f = 18. 0 ? 0. 125 sSpeed (v) =f? = 8. 0 x 0. 40 = 3. 20 m s-1A wave which results in a net transfer of energy from one place to another is known as a progressive wave. | | Intensity {of a wave}: is defined as the rate of energy flow per unit time {power} per unit cross-sectional area perpendicular to the direction of wave propagation. Intensity = Power / Area = Energy / (Time x Area) For a point source (which would emit spherical wavefronts), Intensity = (? m? 2xo2) / (t x 4? r2) where x0: amplitude ; r: distance from the point source. Therefore, I ? xo2 / r2 (Pt Source) For all wave sources, I ? (Amplitude)2 Transverse wave: A wave in which the oscillations of the wave particles {NOT: movement} are perpendicular to the direction of the propagation of the wave. Longitudinal wave: A wave in which the oscillations of the wave particles are parallel to the direction of the propagation of the wave. Polarisation is said to occur when oscillations are in one direction in a plane, {NOT just â€Å"in one direction†} normal to the direction of propagation. {Only transverse waves can be polarized; longitudinal waves can’t. }Example 2: The following stationary wave pattern is obtained using a C. R. O. whose screen is graduated in centimetre squares. Given that the time-base is adjusted such that 1 unit on the horizontal axis of the screen corresponds to a time of 1. 0 ms, find the period and frequency of the wave. Period, T = (4 units) x 1. 0 = 4. 0 ms = 4. 0 x 10-3 sf = 1 / T = 14 x 10-3 250 Hz| | Superposition Principle of Superposition: When two or more waves of the same type meet at a point, the resultant displacement of the waves is equal to the vector sum of their individual displacements at that point. Stretched String A horizontal rope with one end fixed and another attached to a vertical oscillator. Stationary waves will be produced by the direct and reflected w aves in the string. Or we can have the string stopped at one end with a pulley as shown below. Microwaves A microwave emitter placed a distance away from a metal plate that reflects the emitted wave. By moving a detector along the path of the wave, the nodes and antinodes could be detected. Air column A tuning fork held at the mouth of a open tube projects a sound wave into the column of air in the tube. The length of the tube can be changed by varying the water level. At certain lengths of the tube, the air column resonates with the tuning fork. This is due to the formation of stationary waves by the incident and reflected sound waves at the water surface. Stationary (Standing) Wave) is one * whose waveform/wave profile does not advance {move}, where there is no net transport of energy, and * where the positions of antinodes and nodes do not change (with time). A stationary wave is formed when two progressive waves of the same frequency, amplitude and speed, travelling in opposite directions are superposed. {Assume boundary conditions are met} | Stationary waves| Stationary Waves Progressive Waves| Amplitude| Varies from maximum at the anti-nodes to zero at the nodes. | Same for all particles in the wave (provided no energy is lost). | Wavelength| Twice the distance between a pair of adjacent nodes or anti-nodes. The distance between two consecutive points on a wave, that are in phase. | Phase| Particles in the same segment/ between 2 adjacent nodes, are in phase. Particles in adjacent segments are in anti-phase. | All particles within one wavelength have different phases. | Wave Profile| The wave profile does not advance. | The wave profile advances. | Energy| No energy is transported by the wave. | Energy is transported in the direction of the wave. | Node is a region of destructive superposition where the waves always meet out of phase by ? radians. Hence displacement here is permanently zero {or minimum}. Antinode is a region of constructive superposition where the waves always meet in phase. Hence a particle here vibrates with maximum amplitude {but it is NOT a pt with a permanent large displacement! } Dist between 2 successive nodes / antinodes = ? / 2 Max pressure change occurs at the nodes {NOT the antinodes} because every node changes fr being a pt of compression to become a pt of rarefaction {half a period later} Diffraction: refers to the spreading {or bending} of waves when they pass through an opening {gap}, or round an obstacle (into the â€Å"shadow† region). Illustrate with diag} For significant diffraction to occur, the size of the gap ? ? of the wave For a diffraction grating, d sin ? = n ? , d = dist between successive slits {grating spacing} = reciprocal of number of lines per metre When a â€Å"white light† passes through a diffraction grating, for each order of diffraction, a longer wavelength {red} diffracts more than a shorter wavelength {violet} {as sin ? ? ? }. Diffraction refers to the spreading of waves as they pass through a narrow slit or near an obstacle. For diffraction to occur, the size of the gap should approximately be equal to the wavelength of the wave. Coherent waves: Waves having a constant phase difference {not: zero phase difference / in phase} Interference may be described as the superposition of waves from 2 coherent sources. For an observable / well-defined interference pattern, the waves must be coherent, have about the same amplitude, be unpolarised or polarised in the same direction, ; be of the same type. Two-source interference using: 1. Water Waves Interference patterns could be observed when two dippers are attached to the vibrator of the ripple tank. The ripples produce constructive and destructive interference. The dippers are coherent sources because they are fixed to the same vibrator. 2. Microwaves Microwave emitted from a transmitter through 2 slits on a metal plate would also produce interference patterns. By moving a detector on the opposite side of the metal plate, a series of rise and fall in amplitude of the wave would be registered. 3. Light Waves (Young? s double slit experiment) Since light is emitted from a bulb randomly, the way to obtain two coherent light sources is by splitting light from a single slit. The 2 beams from the double slit would then interfere with each other, creating a pattern of alternate bright and dark fringes (or high and low intensities) at regular intervals, which is also known as our interference pattern. Condition for Constructive Interference at a pt P: Phase difference of the 2 waves at P = 0 {or 2? , 4? , etc} Thus, with 2 in-phase sources, * implies path difference = n? ; with 2 antiphase sources: path difference = (n + ? )? Condition for Destructive Interference at a pt P: Phase difference of the 2 waves at P = ? { or 3? , 5? , etc } With 2 in-phase sources, + implies path difference = (n+ ? ), with 2 antiphase sources: path difference = n ? Fringe separation x = ? D / a, if a;;D {applies only to Young’s Double Slit interference of light, ie, NOT for microwaves, sound waves, water waves} Phase difference betw the 2 waves at any pt X {betw the central 1st maxima) is (approx) proportional to the dist of X from the central maxima. Using 2 sources of equal amplitude x0, the resultant amplitude of a bright fringe would be doubled {2Ãâ€"0}, the resultant intensity increases by 4 times {not 2 times}. { IResultant ? (2 x0)2 } Electric Fields Electric field strength / intensity at a point is defined as the force per unit positive charge acting at that point {a vector; Unit: N C-1 or V m-1} E = F / q F = qE * The electric force on a positive charge in an electric field is in the direction of E, while * The electric force on a negative charge is opposite to the direction of E. * Hence a +ve charge placed in an electric field will accelerate in the direction of E and gain KE { simultaneously lose EPE}, while a negative charge caused to move (projected) in the direction of E will decelerate, ie lose KE, { gain EPE}. Representation of electric fields by field lines | | | | | Coulomb’s law: The (mutual) electric force F acting between 2 point charges Q1 and Q2 separated by a distance r is given by: F = Q1Q2 / 4 or2 where ? 0: permittivity of free space or, the (mutual) electric force between two point charges is proportional to the product of their charges ; inversely proportional to the square of their separation. Exa mple 1: Two positive charges, each 4. 18 ? C, and a negative charge, -6. 36 ? C, are fixed at the vertices of an equilateral triangle of side 13. 0 cm. Find the electrostatic force on the negative charge. | F = Q1Q2 / 4 or2= (8. 99 x 109) [(4. 18 x 10-6)(6. 6 x 10-6) / (13. 0 x 10-2)2]= 14. 1 N (Note: negative sign for -6. 36 ? C has been ignored in the calculation)FR = 2 x Fcos300= 24. 4 N, vertically upwards| Electric field strength due to a Point Charge Q : E = Q / 4 or2 {NB: Do NOT substitute a negative Q with its negative sign in calculations! } Example 2: In the figure below, determine the point (other than at infinity) at which the total electric field strength is zero. From the diagram, it can be observed that the point where E is zero lies on a straight line where the charges lie, to the left of the -2. 5 ? C charge. Let this point be a distance r from the left charge. Since the total electric field strength is zero, E6? = E-2? [6? / (1 + r)2] / 4 or2 = [2. 5? / r2] / 4 or2 (Note: negative sign for -2. 5 ? C has been ignored here) 6 / (1 + r)2 = 2. 5 / r2 v(6r) = 2. 5 (1 + r) r = 1. 82 m The point lies on a straight line where the charges lie, 1. 82 m to the left of the -2. 5 ? C charge. Uniform electric field between 2 Charged Parallel Plates: E = Vd, d: perpendicular dist between the plates, V: potential difference between plates Path of charge moving at 90 ° to electric field: parabolic. Beyond the pt where it exits the field, the path is a straight line, at a tangent to the parabola at exit. Example 3: An electron (m = 9. 11 x 10-31 kg; q = -1. 6 x 10-19 C) moving with a speed of 1. 5 x 107 ms-1, enters a region between 2 parallel plates, which are 20 mm apart and 60 mm long. The top plate is at a potential of 80 V relative to the lower plate. Determine the angle through which the electron has been deflected as a result of passing through the plates. Time taken for the electron to travel 60 mm horizontally = Distance / Speed = 60 x 10-3 / 1. 5 x 107 = 4 x 10-9 s E = V / d = 80 / 20 x 10-3 = 4000 V m-1 a = F / m = eE / m = (1. 6 x 10-19)(4000) / (9. 1 x 10-31) = 7. 0 x 1014 ms-2 vy = uy + at = 0 + (7. x 1014)( 4 x 10-9) = 2. 8 x 106 ms-1 tan ? = vy / vx = 2. 8 x 106 / 1. 5 x 107 = 0. 187 Therefore ? = 10. 6 ° Effect of a uniform electric field on the motion of charged particles * Equipotential surface: a surface where the electric potential is constant * Potential gradient = 0, ie E along surface = 0 } * Hence no work is done when a charge is moved along this surface. { W=QV, V=0 } * Electric field lines must meet this surface at right angles. * {If the field lines are not at 90 ° to it, it would imply that there is a non-zero component of E along the surface. This would contradict the fact that E along an equipotential = 0. Electric potential at a point: is defined as the work done in moving a unit positive charge from infinity to that point, { a scalar; unit: V } ie V = W / Q The electric potential at infinity is defined as zero. At any other point, it may be positive or negative depending on the sign of Q that sets up the field. {Contrast gravitational potential. } Relation between E and V: E = – dV / dr i. e. The electric field strength at a pt is numerically equal to the potential gradient at that pt. NB: Electric field lines point in direction of decreasing potential {ie from high to low pot}. Electric potential energy U of a charge Q at a pt where the potential is V: U = QV Work done W on a charge Q in moving it across a pd ? V: W = Q ? V Electric Potential due to a point charge Q : V = Q / 4 or {NB: Substitute Q with its sign} Electromagnetism When a conductor carrying a current is placed in a magnetic field, it experiences a magnetic force. The figure above shows a wire of length L carrying a current I and lying in a magnetic field of flux density B. Suppose the angle between the current I and the field B is ? , the magnitude of the force F on the conductor is iven by F = BILsin? The direction of the force can be found using Fleming? s Left Hand Rule (see figure above). Note that the force is always perpendicular to the plane containing both the current I and the magnetic field B. * If the wire is parallel to the field lines, then ? = 0 °, and F = 0. (No magnetic force acts on the wire) * If the wire is at right angles to the field lines, then ? = 90 °, and the magn etic force acting on the wire would be maximum (F = BIL) Example The 3 diagrams below each show a magnetic field of flux density 2 T that lies in the plane of the page. In each case, a current I of 10 A is directed as shown. Use Fleming’s Left Hand Rule to predict the directions of the forces and work out the magnitude of the forces on a 0. 5 m length of wire that carries the current. (Assume the horizontal is the current) | | | F = BIL sin? = 2 x 10 x 0. 5 x sin90 = 10 N| F = BIL sin? = 2 x 10 x 0. 5 x sin60 = 8. 66 N| F = BIL sin ? = 2 x 10 x 0. 5 x sin180 = 0 N| Magnetic flux density B is defined as the force acting per unit current in a wire of unit length at right-angles to the field B = F / ILsin ? F = B I L sin ? {? Angle between the B and L} {NB: write down the above defining equation define each symbol if you’re not able to give the â€Å"statement form†. } Direction of the magnetic force is always perpendicular to the plane containing the current I and B {even if ? ? 0} The Tesla is defined as the magnetic flux density of a magnetic field that causes a force of one newton to act on a current of one ampere in a wire o f length one metre which is perpendicular to the magnetic field. By the Principle of moments, Clockwise moments = Anticlockwise moments mg †¢ x = F †¢ y = BILsin90 †¢ y B = mgx / ILy Example A 100-turn rectangular coil 6. 0 cm by 4. 0 cm is pivoted about a horizontal axis as shown below. A horizontal uniform magnetic field of direction perpendicular to the axis of the coil passes through the coil. Initially, no mass is placed on the pan and the arm is kept horizontal by adjusting the counter-weight. When a current of 0. 50 A flows through the coil, equilibrium is restored by placing a 50 mg mass on the pan, 8. 0 cm from the pivot. Determine the magnitude of the magnetic flux density and the direction of the current in the coil. Taking moments about the pivot, sum of Anti-clockwise moments = Clockwise moment (2 x n)(FB) x P = W x Q (2 x n)(B I L) x P = m g x Q, where n: no. of wires on each side of the coil (2 x 100)(B x 0. 5 x 0. 06) x 0. 02 = 50 x 10 How to cite Physics Notes, Papers