10. [2p] A small conducting spherical shell with inne radius a and outer radius

Question

10. [2p] A small conducting spherical shell with inne radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius e and oute radius d. The inmer shell has a total charge -2q and the outer shell has a total charge +3g. Which of the following statements aere(Give ALL correct answers, ie., B. AC, BCD... A) The total charge on the inner surface of the large shell is zero. B) The electric field in the region greater than d is zero. C) The radial component of the electric field in the region e less than less than d is given by -2q(4 pi epsilong 1). D) The total charge o the outer surface of the small shell is -2q E) The radial component of heelectric field in the regiotns tha a is given by +lq(4 pi epsilono F) The total charge on the inner surface of the small shllis -2q G) The total charge on the outer surface of the large she is+1q. Submit All Answars 11. [2pt] A 1.32uC charge is placed ar the center of a conducting spherical shell, and a total charge of 6.90?C ts placed on the shell itself. Calculate the total charge on the outer surface of the conductor. Submit All Answers 12. [2p] An solated conductor of a batary shape has a net chage of 1 2x 10" C. Inside the conductons a cavity within which s a pont charge q-70× 10 6 C. what the charge on the cavity wall ? An Submit All Answars 13. [2pt] In the above situation, what is the charge on the outer surface of the conductor? Answer: Submit All Answers

Explanation / Answer

given , conducting spherical shell,

inner radius = a

outer radius = b

qin = -2q

larger conducting spherical shell

inner radius = c

outer radius = d

qo = 3 q

total charge on inner surface will be 0 as all charge will reside on the outher surfac eon the conductor

electric field r > 0 is non zero as net charge is non zero

radial electric field in c < r < d is 0 ( inside the conductor)

Er, r < a = 0 ( from gauss' law)

total charge on outer surface of lager shell is 1 q ( as inner surface will have + 2q to have 0 field inside the conductor)

hecne only option G) is correct

11. q = -1.32 uC

Q = 6.9 uC

hence charge on outer surgface = Q + q = 5.58 uC

12. Q = 1.2*10^-5 C

q = 7*10^-6 C

hence

charge on cavity wall = -7*10^-6 C

13. charg eon outer surface = Q + q = 19*10^-6 C

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