Question

A solid sphere of radius R has a charge Q distributed in its volume with a charge density $\rho=kr^{a}$, where k and a are constants and r is the distance from its centre. If the electric field at $r=\frac{R}{2}\ is\ \frac{1}{8}$ times that at r = R, find the value of a.

Answer 1

Here is your answer ⤵⤵⤵

The total charge is Q. Hence if we integrate the charge we shall get Q

Let us consider a thin spherical shell of thickness dr it'vol is 4∏r² dr and charge will be kra times of this hence From Gauss's law we get the value of the field at r=R/2 is

E 4∏(R/2)²=∫4∏kar³ dr=∏ka(R/2)^4

E=kaR²/16ε...............(1)

again the field at r=R is =Q/4∏R²ε...............(2)

The ratio is 1:8 hence we get

a=Q/(2∏kR^4)

HOPE IT HELPS YOU ☺☺ !!!

Answer 2

Hii Mate.....

Here is your answer ⤵⤵

The total charge is Q. Hence if we integrate the charge we shall get Q

Let us consider a thin spherical shell of thickness dr it'vol is 4∏r² dr and charge will be kra times of this hence From Gauss's law we get the value of the field at r=R/2 is

E 4∏(R/2)²=∫4∏kar³ dr=∏ka(R/2)^4

E=kaR²/16ε...............(1)

again the field at r=R is =Q/4∏R²ε...............(2)

The ratio is 1:8 hence we get

a=Q/(2∏kR^4)

HOPE IT HELPS YOU !!