Question

A spherical shell of radius Ra carries a total charge qa uniformly distributed on its surface. A larger spherical shell of radius Rb is concentric with the first and carries a charge qb uniformly distributed on its surface. (a) Use Gauss’ law to find E~ in all regions (do not forget the inside of the smaller shell!). (b) If qa=6nC, what should qb be for the electric field to be zero for r > Rb? (c) Sketch the electric-field lines for the situation in part (b).

Answer 1

Answer:

Explanation:

Part a)

For a position of point inside the inner shell we can use Gauss law as

$\int E. dA = \frac{q}{\epsilon_0}$

now here we know that enclosed charge in the inner shell is ZERO

so we have

$\int E. dA = 0$

$E = 0$

Now for the position between two shells

$r_a< r< r_b$

again by Gauss law

$\int E. dA = \frac{q}{\epsilon_0}$

now here we know that enclosed charge between two shells is given as

$q = q_a$

so we have

$\int E. dA = \frac{q_a}{\epsilon_0}$

$E = \frac{q_a}{4\pi \epsilon_0 r^2}$

Now for position outside the shell we will have

$r > r_b$

again by Gauss law

$\int E. dA = \frac{q}{\epsilon_0}$

now here we know that enclosed charge given as

$q = q_a + q_b$

so we have

$\int E. dA = \frac{q_a + q_b}{\epsilon_0}$

$E = \frac{q_a + q_b}{4\pi \epsilon_0 r^2}$

Part b)

If outside the shell net electric field is zero

then we can say

$q_a + q_b = 0$

$q_a = 6 nC$

$q_b = - 6nC$

Part c)