16. (IS points) A point charge O is located at the center of a thick conducting

Question

16. (IS points) A point charge O is located at the center of a thick conducting ID: A radius a. The shell has zero total charge. A thin spherical shell of non-conducting surface charge density surrounds the conductor. The net charge on the non-conductor sc spherical shell with inner radius a and outer material radius ay with uniform NC a2 4 cm 43- 9 em 0 a) Calculate the radial component of the total electric field at a radius 15 cm from the center of symmetry. (In the answer list below, a plus sign denotes a field that points radially outward, a minus sign the opposite direction.) a. E +7.20x10 V/m c. E = +360× 10°V/m e. E-5.60x 10 V/m b) Calculatethe surface charge density on the outer surface of the conducting shell. d.o -298 C/m2 c) Calculate the electric potential at the outer surface of the conductor, given that the potential at infinity is z a. pe=-674 x 105 V b. pe=-5.20×102 v

Explanation / Answer

16. from the given question

charge Q is located in the center of a thick conducting spherical shell, radii a1 and a2

0 charge on shell

non conducting shell, radius a3, uniform surface charge density sigma

sigma*4*pi*a3^2 = QNC

a1 = 3 cm

a2 = 4 cm

a3 = 9 cm

Q = -6 uC

QNC = 18 uC

hence

a. at r = 15 cm

Efrom gauss' law

E(r)*4*pi*r^2 = (Q + QNC)/epsilon

E = k(Q + QNC)/r^2 = 4789333.333333 V/m

option b.

b. on the outer surface of the conducting shell, charge density = sigma2

hence from gauss' law

-sigma2*4*pi*a2^2 - 6*10^-6 = 0

sigma2 = -2.9841*10^-4 C/m^2

option d.

c. electric potential at outer surface of the conductor = V

V = kQ/a2 = -1347000 V=-13.47 *10^5 V

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