A spherically symmetric charged distribution has a charged density given by ρ = a / r, where a is a constant. Find the field within the sphere as a function of r. [

Question

asked Mar 28, 2020 100k views

1 Answer

Hui Zheng · Answer 1 · 2020-04-01T05:46:54+0000

Answer:

Electric field,
E = (a)/(2epsilon)

Given:

charge density,
$\rho = (a)/(r)$

a = constant

Solution:

Area of sphere =
$4\pi r_(2) dr$

Small charge dq on a sphere of thickness r is:

dq =
$\rho * A* dr$

dq =
$4\rho \pi r_(2) dr$ (1)

Integrating both the sides, we get:

$dq = 4\rho \pi r_(2) dr$

$\int dq = \int (a)/(r)* 4\pi r^(2) dr$

$q(r)= 2\pi r^(2)a - 0 = 2\pi r^(2)a$

Therefore, the total charge enclosed by a sphere of radius r is
$q(r)= 2\pi r^(2)a$

Now, by Gauss Law, flux emerging out of the surface is equal to
$(1)/(\epsilon_(o))$ times the total charge enclosed:

$A* E = (q)/(\epsilon_(o))$

$4\pi r^(2)* E = \frac{}{\epsilon_(o)}$

$E = (2\pi r^(2)a)/(\epsilon_(o))$

E = (a)/(2epsilon)

where

E = electric field