Physics: Aspherically 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. [

sorry can t read, cuz too small the words.Explanation:...

Aspherically symmetric charged distribution has a

dondre54

31.12.2021

Electric field, $E = \frac{a}{2epsilon}$

Given:

charge density, $\rho = \frac{a}{r}$

a = constant

Solution:

Area of sphere = $4\pi r_{2} dr$

Small charge dq on a sphere of thickness r is:

dq = $\rho \times A\times dr$

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

Integrating both the sides, we get:

$dq = 4\rho \pi r_{2} dr$

$\int dq = \int \frac{a}{r}\times 4\pi r^{2} dr$

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

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 $\frac{1}{\epsilon_{o}}$ times the total charge enclosed:

$A\times E = \frac{q}{\epsilon_{o}}$

$4\pi r^{2}\times E = \frac{}{\epsilon_{o}}$

$E = \frac{2\pi r^{2}a}{\epsilon_{o}}$

$E = \frac{a}{2epsilon}$

where

E = electric field

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