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CK Cheung ELECTROSTATICS Glass ( + ) Fur ( + ) + + Rubber ( - ) Ebonite ( - ) Experimental facts: 1/ There are 2 , & only 2, kinds of electric charge 2/ Like charges repel, unlike charges attract. 3/ Fundamental charge = e- = p+ = 1.6 x 10-19 Coulomb (C). Coulomb’s Law (For point charges) +Q F r +Q F By experiment, F 1 r2 F F Q1Q2 F = k ( Q1Q2 ) r2 Q1Q2 r2 where k = 9x109 Nm2/C2 vacuum) In general, Q1Q2 F= 4 0 r 2 1 (0 = 8.85418x10-12 C2/Nm2) Called permittivity constant 1 CK Cheung Properties of Coulomb Force: 1) Obey Superposition Principle ..\..\powerpoint\couloml,s law\superposition of forces.ppt +Q r1 1 2) 3) +Q -Q F1 F2 r2 1 A conservative force Acts along the line joining the 2 point charges. e.g. 2 identical point charges repel each other with a force 0.1 mN. They are moved 5 mm further apart, and the repulsive force is reduced to 25 N. 1. 2. How far apart were they originally? What was the size of the charges. 1987-I-30 +Q 5 mm 5.3x10-10C -Q W X Z Y q -Q Three charges +Q, -Q and -Q are fixed at the corners W, X and Y respectively of a square as shown. A fourth charge, q, is fixed at Z, after which the charge at X experiences a NET electrostatic force indicated by the arrow. q is equal to A. +Q B + 2Q C + 4Q D + E +2 Q 2 2 Q 2 CK Cheung Electric Field Q +q r P Force transmitted by what ? field concept An electric field is said to exist at a point P if a force is exerted on a stationary test charge (+) placed at that point. Definition: The electric field strength, E, at P is: E= F q NC-1 Qq 4 0 r 2 E= q 1 Hence, E= Q 4 0 r 2 1 1 (for radial field only) r2 Note: 1/ Actually, q must not carry too many charge (q 0 ), as not to affect the charge distribution of Q. (Why do we not define E as numerically equal to the force acting on a charge of one coulomb?) 2/ F = qE 3/ Apply to region outside any spherical conductors (~ point charge). 3 CK Cheung Representation of E Use field line concept: 1) The tangent to a line of force at any point gives the direction of E at that point. E shows the initial direction of motion of a resting positive charge. 2) The number of field lines drawn per unit cross-sectional area is proportional to the magnitude of E Strong E weak E Note: = E A = EAcos Called electric flux 4 CK Cheung Properties of electric field lines 1/ They begin from +ve charge & end on –ve charge - + 2/ They cannot cross (otherwise, direction of E at a point is not unique) ..\..\powerpoint\E-field line property.ppt 3/ They cut equipotential surfaces (i.e. surface of equal potential e.g. conductor) at 900 + + + + + + + + + + + conductor + + + + + + + + 4/ The electric potential is decreasing along the electric field line direction. 5 CK Cheung 1984-I-48. Y Z X The diagram shows a pattern of electric field lines in which X, Y and Z are points marked on one of the field lines. It would be correct to say that (1) (2) X is at a higher potential than Z. a negative charge placed at Z would accelerate to the left along the tangent to the field line at Z. (3) the force exerted on a charge at Y would be greater than if the charge were placed at X. A. (1), (2) and (3) B. (1) and (2) only C. (2) and (3) only D. (1) only E. (3) only 1989-I-30 IV X III II E I Y E A particle carrying a negative charge is free to move in a uniform electric field E. If the particle starts with a certain velocity from point X, which of the paths shown could represent the route which the particle would follow from X to Y ? A. I and II only B. I and IV only C. II and III only D. III and IV only E. I, II, III and IV 1999-II-21(AL) Which of the following statements about electric field lines is incorrect ? A. They are closest where the field is strongest. 6 CK Cheung B. They are always perpendicular to equipotential lines. C. They always point from high electric potential to low electric potential. D. Work has to be done in moving an electron along the direction of a field line. E. They tend to attract one another. 7 CK Cheung Relative magnitude of gravitational & electrostatic forces. + R Fe = 1 Q1Q2 40 R 2 (9 x10 9 )(1.6 x10 19 ) 2 = R2 Fg = GMeMp (6.67 x10 11 )(9.11x10 31 )(1.6 x10 27 ) = R2 R2 Fe Fg 10 39 ∴ In general , Fe >> Fg ..\..\powerpoint\This demonstrates that Coulomb force ( not gravitational.ppt This demonstrates that Coulomb force ( not gravitational forces ) are responsible for binding : Electrons to nuclei to form atoms Atoms to atoms to form molecules Molecules to molecules to form solids and liquids 8 CK Cheung E – field for a group of point charges ..\..\powerpoint\E-field due to a system of charges.ppt P Calculate E-field due to individual charge at the given point as if it were the only charge present ( i.e. superposition principle ) and then add vectorially. 9 CK Cheung e.g. Refer to the following diagram, in which r >> 2a. Find, in magnitude, EL ? ET T r -q 2a L +q r Q 1 1 ( ) EL = 2 4 (r a) (r a) 2 ET = = Q 1 4 (( a r ) ) 2 2 4 1 2 2 Q 1 (cos Θ) x 2 = 4 2 a 1 2 (( a r ) ) (a r ) 2 2 2 2 2 1 2 x2 3 Qa (a r ) 2 2 2 Q 1 a a Q 1 4a ( ) = 1 1 = 2 4 r 2 r r 4 r r 2 Hence, EL/ET = 2 3 2 = 2 Qa a2 2 ( 1 ) 4 r 3 r2 0 10 CK Cheung e.g. A point charge +Q is placed a distance R from an infinite conducting thin sheet (earthed) on which it induces a charge -Q. a/ draw the electric field lines for this arrangement. ..\..\powerpoint\e-field lines.ppt -Q +Q (note: -Q is not evenly distributed on the surface of the conducting sheet) b/ draw the field pattern associated with point charges +Q & -Q separated by a distance of 2R. -Q +Q R c/ compare (a) & (b) , find the attractive forces that the plane & the point charge +Q exert on each other. ..\..\powerpoint\e-field lines2.ppt F= Q2 4 0 (2 R) 2 1 11 CK Cheung Formula: 1/ Infinite conducting thin sheet + Uniform E = 2 0 σ = surface charge density 2/ 2 infinite sheets ..\..\powerpoint\e-field lines3.ppt + + Uniform E = 0 E=0 Uniform E = 12 0 CK Cheung 3/ Infinite conducting wall Uniform E = E=0 0 13 CK Cheung Electric Potential Consider: hill WD = mgh = PE difference between top & bottom g Similiary: B +q If we carry +q from A to B, & work must be done = WAB there exists potential difference between A & B. A Definition: ΔV = VB – VA = W AB Volts ( V ) q Note: 1/ the charge must be in steady speed when moving it from A to B, otherwise WAB ≠ unique value. 2/ If A → , define VA = 0 VB = W B definition of electric potential at a point q 14 CK Cheung Consider: X E Bring +q from X to Y: qE Y W XY VY-VX = = q +q qE d x X q Y ∴ ΔV = - E dx X from maths.: FApply = -qE E=- dV dx Y -ve E points in the direction of decreasing potential. 15 CK Cheung e.g. Find the potential difference between A & B : rA rB FApply +Q +q B W AB VB – VA = = q VB – VA = B ( A 1 Qq )dl 4 r 2 = q E dl B 1 Q A 4 r 2 (- dr) = rB A 1 Q 4 r 2 (- dr) rA Q 1 1 ( ) >0 4 rB rA potential decreases along the direction of E-field lines If r A VA ~ 0 ∴ VB = 1 Q ) 4 rB ( hence, in general: V= Q 4 r 1 Note: 1. V > 0 or V<0 2. For radial field only 3. If +Q is a sphere treat it ~ point charge ( ∵ also radial E-field),the formula holds for r > radius of sphere. 16 CK Cheung Potential due to a group of charges: P VP = V i i = i 1 qi ( ) 4 ri ( algebric sum ) e.g. Q1 Q2 a What is the electric potential at the center of the square? Given: Q1=+1x10-8 C Q2= -2x10-8 C Q3=+3x10-8 C Q4=+2x10-8 C Q4 V= Q3 V i i a=1m Q1 Q2 Q3 Q4 1 1 2 3 2 ( )(10 8 ) 500 V ) = 4 0.7 4 r 1 ( 17 CK Cheung Potential & E-field inside conductors 1/ thin spherical shells + + + + E-field = 0 inside the outer surface because: Current = 0 All charges are at outer surfaces R ∵ E= + + ∴ if E - dV dx =0 dV =0 dx V = constant of position All points are at the same potential inside the outer surface potential at all points inside the shell = surface potential of shell + + + + R + R + ∴ surface potential = Q ) 4 R 1 ( For r R 18 CK Cheung 2/ Solid conducting sphere potential = Q ) 4 R 1 ( Hence: for Shell or sphere: R E= 1 Q 4 r 2 E=0 V= V= 1 Q 4 r 1 Q 4 R 19 CK Cheung e.g. Find the p.d. between the two concentric spherical shells? q Q R r Vr = 1 Q q + 4 R 4 r VR = 1 q 1 Q + 4 R 4 R 1 Vr – VR = q 1 1 ( ) 4 r R Independent of Q 20 CK Cheung e.g. Two charged conducting spheres are connected as shown. In steady state, the charges on them are q1 & q2 σ1 q1 Long wire R1 R2 Prove that 1 R2 2 R1 σ2 q2 Proof: V1 = 1 q1 ( ) 4 R1 ∵ V1 = V2 & 1 q1 ( ) 4 R1 V2 = = 1 q2 ( ) 4 R2 1 q2 ( ) 4 R2 q1 R1 q 2 R2 (q R) q1 R R2 4R12 q R2 ∴ 1 = 1 . 22 = ( 1 )( 22 ) q2 2 R2 R1 q 2 R1 2 4R 2 1 R2 2 R1 point action ( 1 ) R 21 CK Cheung E – field between 2 parallel plates E + - + + + + - +q + + + - X ∵ ΔV = or WD (qE ) X EX q q E= V X For uniform field only potential E-field 0 X/m 22 CK Cheung Potential due to neighboring bodies -q P +Q R Vp = 1 Q ) 4 R ( r Vp = - 1 q ) 4 r ( 23 CK Cheung e.g. Sketch the V-X & E-X graphs both before & after the conducting slab is inserted. (i.e. 4 curves in 2 graphs) + + + + + + + + - ..\..\powerpoint\e-field & potential.ppt V/V E/Vm-1 X/m X/m 24 CK Cheung Electric Potential Energy rA rB FApply +Q +q B E A By def.: ∵ ΔV = VB-VA = W AB q ∴ WAB = q (VB-VA) from above diagram: WAB = q ( 1 Q 1 Q ) 4 rB 4 rA If rA = , take VA = 0 ∴ WB = qVB = 1 qQ 4 rB = electrostatic P.E. of the system Def: We define the electric P.E. of a system of point charges as the work required to assemble the system of charges by bringing them in from an infinite distance. In general: U= 1 ( q1 )( q2 ) ( ) 4 r >0 ( repulsion) Note: U <0 (attraction) 25 CK Cheung e.g. Three charges are held fixed as shown. What is the electrical P.E. of the system? Take q = 1x10-7 C, a = 10 cm. ( hint: use superposition principle) 2 -4q a a 1 +q a 3 +2q Soln.: Use superposition principle: U = U12 + U23 + U13 = 1 ( q)( 4q) ( q)( 2q) (4q)( 2q) 4 a a a = -9x10-3 J Note: U <0, +9x10-3 J of work must be done to separate the structure. 26 CK Cheung e.g. A particle of charge +Q is fixed at P. A second particle of mass m and charge –q moves at constant speed, in a circle of radius R1, centered at P. Find the work that must be done by an external agent on the second particle in order to increase the radius of the circle of motion to R2. R2 -q m +Q R1 At r = R1: F= 1 Qq mv 2 = 4 R12 R1 1 Qq mv 2 = 8 R1 2 1 Q(q) P.E. = 4 R1 K.E. = 1 ∴ Total energy , E1, = K.E. + P.E. = - Qq 8 R1 Similarly, Total energy, E2, at r = R2 = - Qq 8 R 2 W.D. = E2 – E1 = 1 Qq 1 1 ( ) 8 R1 R2 27 CK Cheung Capacitance Consider any pair of isolated conductors +q -q V ∵ V q ∴ q V = constant, called capacitance of the system, C or C= q V units of C : CV-1 Farad ( F ) 1 uF = 10 –6 F 28 CK Cheung Calculating Capacitance 1/ Spherical shells: -Q +Q r R Vr = 1 Q 1 Q + 4 R 4 r VR = 1 Q 1 Q + 4 R 4 R Vr – VR = ∴ C= Q 1 1 ( ) 4 r R Q = V 4 1 1 ( ) r R Note: 1. C depends on 2. If R= Isolated sphere C = 4 r 3. If C r= = 1 F, geometrical factors 1 4 only = 9x109 m !! ( c.f. RE ) 29 CK Cheung 2/ Parallel plate type Area = A +q E d -q ∵ E= q q = EA A Also, V = E(d) ∴ C= q EA V Ed C= ε A d 30 CK Cheung e.g. a/ A parallel-plate capacitor whose plates have an area 1 m2 and which are separated by 2 mm is connected across the terminals of a 100V battery. Find: Area=1m2 100 V 0.002m 1/ the electric field between the plates. ∵ V = Ed ∴ E= 100 5 x10 4 NC-1 0.002 2/ the magnitude of the charge density on the plates ∵ E= E (5x10 4 )(8.85x10 12 ) 4.43x10 7 C m-2 3/ the charge on either plate Q = A 4.43x10 7 (1) = 4.43x10-7 C 4/ the capacitance of the system Q 4.43x10 7 4.43x10 9 F ∵ C= V 100 31 CK Cheung b/ A conducting plate 1 mm thick is now inserted between the plates, the battery being still connected, Find: 1/ the electric field inside the conducting plate E=0 inside conductors 2/ the electric field in the remaining air space ∵ V = E d 100 = E(0.002-0.001) E = 105 Vm-1 3/ the induced surface charge density on the faces of the inserted plates. ∵ each plate induces equal charges on the surface of the conducting plate. ∴ E= E (10 5 )(8.85x10 12 ) 8.85x10 7 Cm-2 Q = A (8.85 x10 7 )(1) 8.85 x10 7 C 4/ the capacitance of the system Q 8.85 x10 7 ∵ C= 8.85x10-9 F C increases V 100 C/ If the conducting slab is inserted after the capactor is disconnected from the 100V battery, repeat part (b) above 32 CK Cheung Capacitors in parallel q1 V C1 q2 C2 q1 = C1V q2 = C2V ∴ total charge Q = q1 + q2 = ( C1+C2)V Q C1 C 2 V = CEQUIVALENT = Ceq ∴ The system is equivalent to a single capacitor of capacitance: Ceq = C1+C2+………….. Note: Charge on Ceq = q1+q2+……….. 33 CK Cheung Capacitors in series +q +q -q -q C2 C1 + V Total charge stored in the system = q - ∵ V = V1 + V2 = ∴ q V q q 1 1 q( ) C1 C 2 C1 C 2 1 1 1 ( ) C1 C 2 = Ceq Hence , the system is equivalent to a single capacitor of capacitance: 1 1 1 .................. Ceq C1 C 2 Note: 1/ Qeq = qi (charge on any one ) 2/ charge stored in each capacitor is the same 34 CK Cheung C1=1 F e.g. C3=3 F S2 C2=2 F C4=4 F 12 V S1 Find the charge on each capacitor when: a/ S1 is closed b/ S2 is also closed solution: a/ Given circuit is equivalent to : C C* 12 V 1 1 1 3 C F C C1 C3 4 1 1 1 4 C* F * C2 C4 3 C ∵ q = CV = ( also: 3 F)(12) = 9x10-6 4 q* = C*V = ( C = q1 =q3 4 F)(12) = 16x10-6 3 C = q2 =q4 35 CK Cheung b/ C1=1 F C2=2 F C3=3 F C4=4 F 12 V Hence, given circuit is equivalent to: C1=1 F V1 C2=2 F C3=3 F V2 C4=4 F 12 V ∵ q1 + q 2 = q 3 + q 4 V1 + 2V1 = 3V2 + 4V2 3V1 = 7 V2------------- (1) Also, V1 + V2 = 12----------(2) V1 = 8.4 V & V2 = 3.6 V q1 = C1V1; q2 = C2V1 ; q3 = C3V2; q4 = C4V2 36 CK Cheung 87’ MC 28 e.g. A 3μF 4μF C 1200 V 450 V 1μF 2μF D B If after some time, the voltage across CD suddenly jumps to 600V, which capacitor(s) has been shorted ( i.e. become a conductor)? 1/ 2 μF shorted: VCD = 0 2/ 1 μF shorted charges on 4 μF & 2 μF will neutralize each other VCD = 0 3/ 3 μF shorted: If VCD = 600 V, VAC also = 600 V ! impossible ! ∵ Q2 = Q4 ( series) but C2 C4 V2 V4 4/ 4 μF shorted: given circuit is equivalent to: answer. 3μF 600V 1200V 3μF 600V 37 CK Cheung Energy stored in a capacitor -q +q V C Vo Let q = charge at any instant before the capacitor is saturated (q & V increase, until q = Q & V = Vo ∵ WD = qV dw = V(dq) = Q ∴ WD = but C= q cons tan t ) V q (dq ) C q (dw) ( C )dq 0 1 Q2 WD = 2 C or WD = 1 CVo2 2 energy stored in the capacitor in the form of E-field 38 CK Cheung e.g. S Qo Vo V C1 C2 C1 is charged to V0 and Qo and then connected to a C2 ( uncharged ) by closing S. 1/ what is the final p.d. across the combination? Qo = Q1 + Q2 C1V0 = C1V + C2V C1 V = V0 ( ) C1 C 2 This suggest a way to measure an unknown C2(e.g.) in terms of a known one. 2/ what is the stored energy before & after S is closed? Initial energy = U0 = 1 C1V02 2 Final energy = U f = 1 1 1 C1V 2 + C 2V 2 = V 2 (C1 C 2 ) 2 2 2 = VC 1 (C1 C 2 )( 0 1 ) 2 2 C1 C 2 = ( C1 )U 0 C1 C2 U f< U0 the missing energy appears as heat in the connecting wires as the charges move through them. 39 CK Cheung e.g. (a) Two // metal plates, each of area 900 cm2, are mounted 2 mm apart in air. If the 2 metal plates are connected to the terminals of a 1.5 V battery, find 1/ the capacitance of the capacitor, C= 1.5 V A 900 x10 4 8.85 x10 12 d 2 x10 3 ∴ C = 3.98x10-10 F 2/ the energy stored in the capacitor. Energy = 1 1 CV 2 (3.98 x10 10 )(1.5) 2 4.48 x10 10 J 2 2 (b) 1.5 V P x P is a point between the 2 plates and is at a distance x from the lower plate. Draw sketches to show how 1/ the E-field at point P, 2/ the potential V at point P varies with x. 1/ E= V 1.5 750Vm 1 x 2 x10 3 E/Vm-1 750 X/m 2x10 -3 40 CK Cheung 2/ V = Ex linear V/V 1.5 X/m 2x10 -3 (c) with the separation of the plates fixed at 2 mm, the top plate is then shifted sideways so that the overlapping area of the plates is reduced to 600 cm2. Neglecting the edge effects, what is the energy stored in the capacitor if its plates: 2 mm 1/ remain connected to the 1.5 V battery 2/ are disconnected from the battery, while the top plate is being moved? In each case, why is the answer different from that in (a)(ii)? Explain briefly. A f (8.85 x10 12 )(600 x10 4 ) 1/ Cf = ε 2.66 x10 10 F 3 d 2 x10 1 1 Ef = C f V 2 (2.66 x10 10 )(1.5) 2 2.99 x10 10 J 2 2 Lesser energy stored in a system of lesser capacitance with same applied voltage 41 CK Cheung 2/ ∵ disconnected from battery, Q unchanged Qf = Qi Qi = CiVi = (3.98x10-10)(1.5) = Qf = 5.97x10-10 C Ef = 1 Qi2 1 (5.97 x10 10 ) 2 6.7 x10 10 J 10 2 C f 2 (2.66 x10 ) More energy stored in the system due to W.D. by external agent in pulling the top plate. spontaneous 42 CK Cheung The RC ( dc) circuit R a b C E 1/ charging (S a) At any time: q C E=iR+ d di 1 dq (E) = R ( ) + dt C dt dt 0 = R( - RC di 1 (i ) ) + C dt di dt i i - RC i0 (1) -RC ln ( t di dt i 0 i )=t i0 i i0 e hence, t ( ) RC (Where: i 0 = original current at t = 0) At t = 0, i i 0 & q = 0 From (1): i 0 = E R t ( ) E i ( )e RC R 43 CK Cheung i /A i0 E R ~ e-t t/s 0 For the charges q on the capacitor: ∵ i dq dt q= (i)dt = t t E ( ) 0 ( R e RC )dt q = CE ( 1- e ( t ) RC ) q/C q0=CE 0 t/s 44 CK Cheung Definition Capacitative = τ= RC Time constant At t = τ q = q0 ( 1- e ( RC ) RC ) q = q0 ( 1- e 1 ) q = q0 ( 1- 0.37 ) q 0.63 q0 capacitor is charged up to 63 % of its equilibrium (saturated ) value q/C q0=CE 0.63q0 0 τ t/s Note: 1. if t > 3 τ circuit can be assumed to be in steady state 2. for calculation, put t = for steady state 45 CK Cheung e.g. S R C E After how many time constant ( RC ) will the energy stored in the capacitor reach half of its equilibrium value? t ( ) 1 q2 1 RC ∵ Ut = CE ( 1 e ) 2 C 2C 2 t = but, ( ) 1 CE 2 (1 e RC ) 2 2 U = 1 CE 2 2 Ut = U (1 e ( t ) RC 2 ) t ( ) Ut 1 RC 2 = (1 e ) = 2 U t = 1.22 RC 46 CK Cheung 2/ Discharging ( S b ) R a b VC= E At t = 0, VC= i i Q = C Q C C E VC E i0 = R R decreases as t = i - i0 e ( t ) RC increases direction reversed also, VC =E e ( t ) RC and q=Q e ( t ) RC VC/q/ i t/s 47 CK Cheung 89’ 26. An isolated spherical conductor of radius r is charged to a potential V. The total electrical energy stored is A. V²/(80 r). B. V²/(40 r). C. V²/(20 r). D. V² (20 r). E. V² (40 r). E = 1 1 CV 2 = ( 4r) V2 2 2 V² ( 2r). 87’ 29. i /m A 10 5 t /s 0 10 20 30 40 A capacitor C is charged to a certain pd. and then discharged through a resistor R. The variation of the current i with time t is shown in the above graph. Which of the following is/are correct? (1) The time constant of the circuit is about 15 s. (2) The area under the graph is proportional to the energy stored in the capacitor. (3) If the resistance of R is doubled, the current at t = 0 will also be doubled. A. (1), (2) and (3) B. (1) and (2) only C. (2) and (3) only D. (1) only E. (3) only 1/ correct 2/ E = i2 R t = R i 2 dt area under i2 – t curve incorrect 3/ i0 = E ; if R increases i0 decreases R incorrect 91’ essay (6)--------self study 48 CK Cheung 85’ IIB 5. GM tube amplifier R C microammeter 1V 50 s Figure 2 Figure 3 Figure 2 shows a simplified diagram of a ratemeter. The signal from the GM tube is fed into an amplifier. After shaping and amplification, the output signal is a square wave of amplitude 1 V and duration 50 s (see Figure 3). Using an RC circuit, a corresponding current signal is observed in a microammeter. When the GM tube is brought near a radioactive source, a steady pulse train of 100 pulses per second ( T= 0.01 s )is produced and measured. The capacitance C and resistance R are of values 0.5 F and 3 k, respectively. The output impedance of the amplifier is small during the charging of C but large during its discharging. (a) (i) Calculate the time constant for the RC circuit. τ= RC = (3000)(0.5x10-6) = 1.5x10-3 s (ii) A student claims that the capacitor is almost completely discharged during the interval between successive pulses. Justify this claim. 50 μs t T=0.01s t= 0.01 - 5x10-5 = 9.95x10-3s >> τ capacitor is almost completely discharged. 49 CK Cheung (b) Calculate the current through the microammeter. In 1 pulse, q stored = CV = (0.5x10-6)(1) C Current = q x frequency = (0.5x10-6)(100) = 5x10-5 A (c) The original radioactive source is replaced by a weaker one such that, on average, only one pulse per second is produced by the GM tube. In order to have a steady reading on the microammeter, should the time constant of the RC circuit be increased or decreased? Explain your answer. (8 marks) τ should be increased. Whenτ >> interval between pulses p.d. across the capacitor remains more or less constant & thus gives a steady current. 50 CK Cheung e.g. 81’ (4) 4. A student is asked to perform an experiment to demonstrate that the charge on a capacitor is proportional to the p.d. across its terminals. The circuit to be used is shown in Figure 2. A 10 V C S to oscilloscope y-input; timebase off; y-sensitivity set to d.c. 2 V/cm Figure 2 The principle of the experiment is to close switch S and charge the capacitor linearly by continuously adjusting the variable resistor to keep the current in the circuit constant at its initial value. The p.d. across the capacitor will therefore increase linearly with time, and a graph of the deflection of the oscilloscope spot in the y-direction against time will be a straight line. (a) The student decides that it would be most convenient to start with the variable resistor set to its maximum value of 100 k and to use an ammeter which he can maintain at full scale deflection when the switch is closed. What should be the full scale deflection of the meter? i= V 10 10 4 A R 100 x10 3 (b) Having selected his meter, the student considers the deflection of the oscilloscope spot. He decides that he will be able to time a deflection rate of 0.1 cm/s satisfactorily. he choose? What value of C should Voltage changing rate = (2 V cm-1)(0.1 cm s-1 ) = 0.2 V s-1 ∵ V= Q it V i 10 4 0.2 = C C t C C C = 5x10-4 F 51 CK Cheung (c) What will be the value of the variable resistor 10 s after closing the switch S? After 10 s. p.d. R= p.d. across C = ( 0.2 V s-1) (10) = 2 V across R = 10 – 2 = 8 V 8 = 8 x 104 10 4 Ω 52 CK Cheung 86’ IIB 4 4. V S R 12 V C A Figure 3 A student performs an experiment in which a 500 F capacitor C is to be charged at a constant rate, using the circuit in Figure 3. After switch S is closed, the microammeter shows an initial current reading of 120 A. By continuously adjusting the variable resistance R, the charging current is maintained at this magnitude throughout the subsequent charging process. V /V R 12 10 8 6 4 2 0 50 100 150 200 250 300 time / s Figure 4 53 CK Cheung At the same time, the student also notes down, from the voltmeter V, the variation of the p.d. across R (VR) as a function of time. The observed variation is shown in Figure 4. The student then draws the conclusion that the capacitor is leaking. In answering the following questions, you may assume that a leaking capacitor can be regarded as a capacitor with a high resistance R0 (the leakage resistance) connected in parallel. R0 (a) Sketch on Figure 4, the expected variation of VR with time if the capacitor is NOT leaking. V /V R ∵V=VR+ 12 Q C it V = VR + C 10 it VR= 12 - C 8 st. line 6 4 2 0 50 100 150 200 250 300 time / s Figure 4 54 CK Cheung (b) Explain the difference between the two curves, and indicate the reason why the fact that VR approaches a constant value after 200 s suggests a leaking capacitor. Since the capacitor is charged at a constant rate, the voltage across the capacitor will also increase at a constant rate. So, the voltage across the variable resistor will decrease at a constant rate. This supports the curve for a not leaking capacitor to be a st. line. The reason for the voltage across the resistor drops to a constant value is that when the capacitor is charged up (i.e saturated when t ~ 200 s), a steady current will pass through the leaking resistor, voltage across the variable resistor is then steady. (c) (i) Sketch below the variation of charge Q stored in capacitor C as a function of charging time t. (No numerical values to be calculated.) Q T/s 200s (ii) Calculate the final charge stored on capacitor C. Q = CV = (500x10-6)(12-2.4) = 4.8x10-3C (iii) Calculate the leakage resistance R0. (12 marks) V=iR (12-2.4) = (120x10-6)(R) R = 80000 Ω 55