A solid sphere of radius $R$ has a charge ${\cal Q}$ distributed in its volume with a charge density ${\rm{\rho }} = k{r^a},$ where $k\;\;{\rm{and\;}}a{\rm{\;}}$ 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 $a.$

JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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Application +4 | -1

A solid sphere of radius

R

has a charge

{\cal Q}

distributed in its volume with a charge density

\mathrm{\rho } = k{r^a},

where

k\;\;\mathrm{and\;}a\mathrm{\;}

are constants and

r

is the distance from its centre. If the electric field at

r = \frac{R}{2}

\frac{1}{8}

times that at

r = R,

find the value of

a.

View answer and explanation

Correct option: A

Option A (2) is correct because by applying Gauss's law and comparing the electric fields at r = R/2 and r = R, we can solve for the value of a.

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r.

4N

\left( {{\varepsilon _r} = 8} \right)

N

5 \times {10^{ - 2}}