In the figure a solid sphere of radius a = 3.00 cm is concentric with a spherica

Question

In the figure a solid sphere of radius a = 3.00 cm is concentric with a spherical conducting shell of inner radius b = 2.00 a and outer radius c = 2.40a. The sphere has a net uniform charge q_1 = +2.11 fC: the shell has a net charge q_2 = -q_1. What is the magnitude of the electric filed at radial distances (a) r = 0 cm, (b) r = a/2.00, (c) r = a, (d) r = 1.50a, (e) r = 2.30a, and (f) = 3.50 a? What is the net charge on the (g) inner and (h) outer surface of the shell. (a) Number _________ Units (b) Number _________ Units (c) Number _________ Units (d) Number _________ Units (e) Number _________ Units (f) Number _________ Units

Explanation / Answer

a = 3 cm ; b = 2a ; c = 2.4 a; Q = 2.11 fC

We know that for electric field inside a conductor for r <=R is

E = k q r/R^3

In the given case, E = k q r/a^3

a)r = 0 cm

using the above eqn we get:

E = 0

b)r = a/2

E = 9 x 10^9 x 2.11 x 10^-15 x 0.5 x 2 x 10^-2/(2 x 10^-2)^3 = 2.37 x 10^-2 N/C

Hence, E = 2.37 x 10^-2 N/C

c)r = a

E = k q a/a^3 = k q/a^2

E = 9 x 10^9 x 2.11 x 10^-15/0.02^2 = 4.75 x 10^-2 N/C

Hence, E = 4.75 x 10^-2 N/C

d)r = 1.5 a = 1.5 x 2 = 3 cm = 0.03 m

r > a for this case so

E = k q/r^2

E = 9 x 10^9 x 2.11 x 10^-15/0.03^2 = 0.0211 N/C

Hence, E = 2.11 x 10^-2 N/C

e)r = 2.3 a

This comes inside and field inside a conductor is zero

Hence, E = 0

f)r = 3.5 a

total charge enclosed here is zero

E = k q/a^2 = 0

Hence, E = 0 N/C

g) Q(inner) = -2.11 fC

f)Q = -2.11 + 2.11 = 0

Hence, Q(outer) = 0 C

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