Solution to Quiz April 8, 00 Four capacitors are connected as shown below What is the equivalent capacitance of the combination between points a and b? a µf b 50 µf c 0 µf d 5 µf e 34 µf Answer: b (A lazy man s solution: Treat the capacitors as a whole with an equivalent capacitance C 34 Then the total C capacitor C 34 C 34 C +C 34 < C 0µF Only answer b satisfies this!) Solve it step by step: As shown in Fig, C and C 3 are capacitors in series, so their effective capacitance C 3 satisfies + C 3 C C 3 C 3 C C 3 6µF µf µf 4µF C + C 3 6µF + µf Then, the effective capacitor C 3 is parallel to C 4, so we get a new effective capacitor C 34, C 34 C 3 + C 4 4µF + 6µF 0µF which works the same as the combination of C, C 3 and C 4 in this configuration Finally, the equivalent capacitance of the whole system would be C 34, which satisfies C 34 C 34 + C C 34 C C 34 0µF 0µF C + C 34 0µF + 0µF 5µF Four capacitors are connected as shown below If a difference of potential of 8 V is applied between points a and b, what is the potential difference and the charge in the top 6 µf capacitor? a 8 V and 90 µc b 3 V and 54 µc C 0μF C 6μF C 3μF a b Figure : C 46μF
And c 6 V and 36 µc d 3 V and 36 µc e 6 V and 54 µc Answer: c (A lazy man s solution: Check V Q C, C 6µF for the five answer Here only answer c satisfies this!) Solve it step by step: Since C and equivalent capacitors C 34 are in series, they carry the same charge, V Q C, V 34 V C 34 C V 34 Q C 34 V ab V + V 34 C ( ) 34 C34 V 34 + V 34 + V 34 C C V 34 C 34 C + V ab 8V 9V 0µF 0µF + Then, since C 3 is in parallel with C 4, the potential difference between them are the same V 3 V 4 V 34 9V Finally, using the same reasoning as above for finding V 34 when V ab is known, we can find V, the potential difference in the top 6µF capacitor, when V 3 is known And The charge in the top 6µF capacitor is V Q C, V 3 Q C 3 V 3 C C 3 V V 3 V + V 3 V + C ( V + C ) V C 3 C 3 V + C V 3 C 3 + 6µF µf 9V 6V Q C V 6µF 6V 36µC 3 A parallel plate capacitor is made from two square plates, 0cm on a side The plates are spaced 0cm apart and connected to a 50V battery How much energy is stored in the capacitor? a 89 0 8 J b 0 7 J c 35 0 J d 89 0 0 J e 44 0 8 J Answer: e For parallel plate capacitor, its capacitance is C ε 0A d The energy stored in the capacitor is U CV ε 0 A d V (885 0 C ) 0cm 0cm N m (50V ) 0cm 4 45 0 8 J
4 A coaxial cable consists of 40-mm-diamete conductor and an outer conductor of diameter 3cm and negligible thickness If the conductors carry line charge densities ±056nC/m, what is the magnitude of the potential difference between them? a 9 V b V c 4 V d 48 V e 97 V Answer: b The electric field between the inner and outer conductors can be worked out by Gauss s law (taking a Gauss surface: a coaxial cylindrical surface with radius r and length L between the two conductors) EdA q enclosed ε 0 Here, where λ is the charge density of the inner conductor A πr L, q enclosed λl E (r) q enclosed ε 0 A λl ε 0 πr L λ πε 0 r The potential difference between the two conductors is V E dl λ πε 0 ln r outer Here, V λ πε 0 ln r outer router r outer d outer d inner 3cm 40mm 8, λ πε 0 r dr λ ln r r outer r πε inner 0 λ 056nC/m 056 0 9 C/m, 056 0 9 C/m π ( ) ln 8 0 9V 885 0 C N m 5 Assuming that the charges +q, +q and +5q are always fixed, what is the total amount of potential energy lost if bring the charge 3q from infinity to the position shown on the figure Use q 0nC and d m a 47 µj b µj c 0 µj d 0 µj e 333 J Answer: c kq d U tot ( 3q) (V + V + V 3 ) ( kq ( 3q) d + k (q) + k (5q) ) d d ( ( 3) + ) kq + 5 d ( ) 899 0 9 N m C ( 0 0 9 C ) m 3 4 495 0 7 J
( U tot ( 3) + ) + 5 4 5 0 7 J 9 998 0 6 J 0µJ 6 Two protons are 00 0 5 m apart What is the change in electric potential energy if they are brought 00 0 5 m closer together? a 5 0 3 J b 30 0 9 J c 30 0 6 J d 60 0 4 J e 30 0 3 J Answer: a The potential difference between these two places is V kq kq r The change in energy is U q V kq r r (899 0 9 N ) m C ( 6 0 9 C ) 00 0 5 m 00 0 5 m 5 0 3 J 7 A capacitor consists of conducting sphere of radius R surrounded by a concentric conducting shell of radius R What is its capacitance? a ε 0 R /R b ε 0 R R c 4πε 0 R R /(R R ) d 4πε 0 R /R e ε 0 (R R ) Answer: c (A lazy man s solution: Imagine the case that R and R are very large and close to each other So them become a parallel plate capacitor Recall the capacitance of a parallel palter capacitor is C ε 0 A/d In this case A 4πR, (R R or R ), d R R So C ε 0 R / (R R ) Only answer c is in this form!) Solve it step by step: Capacitance is defined as the ratio of charge to potential difference r C Q V Assume the conducting sphere is charged Q, then the potential difference between the two conductors is R V Edl R And the electric field produced by the conducting sphere is equivalent to that produced by a point charge with charge Q located at the center of the sphere E kq r So we have R V R kq dr kq r ( ) ( R R r kq ) R R Q R C ( ) R kq R k (R R ) 4πε R R 0 (R R ) R 4
8 A point charge of +30µC is located at the origin of a coordinate system and a second point charge of 60µC is at x +0m At what point on the x axis is the electrical potential zero? a 05 m b +05 m c +033 m d +075 m e +5 m Answer: c A Lazy man s strategy: The electrical potential at point x is a superposition of the potentials produced by the two point charges V x V + V q + q r r Here, q +30µC, q 60µC So V x 0 V x r r r r r, r are the distances from point x to the location of charge q, x 0 and the location of charge q, x +0m, respectively Only answer c satisfies this! A careful guy s solution: You can go on to work out the location x by solving the equation: r x r x x x x 3 m; or x x m x So, in fact, we have two solutions for the location x The first solution is answer c The second solution is not included here 5