Question

A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric field at a point a distance r from Q? Consider the cases r < a, a < r < b, and r > b.

Answer 1

Answer:

a)$E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

b)E=0

c)$E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a)    r < a

At a radius r ,from gauss theorem

$E.ds=\dfrac{q_i}{\varepsilon _o}$

$E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}$

$E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

b)   a < r < b

$E.ds=\dfrac{q_i}{\varepsilon _o}$

The total induce in this surface = - Q+ Q =0

$E.ds=\dfrac{0}{\varepsilon _o}$

E = 0

c)   r > b

$E.ds=\dfrac{q_i}{\varepsilon _o}$

$E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}$

$E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$