A solid sphere of radius R has a charge Q distributed in its volu… - Vidyadip

MCQ

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=\kappa r^a$, where $\kappa$ 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 $a$.

A
$1$
✓
$2$
C
$3$
D
$4$

✓

Answer

Correct option: B.

$2$

b
Using Gauss's law, we have

$\oint \vec{E} \cdot d \vec{A}=\frac{1}{\in_0} \int(\rho d v)=\frac{1}{\in_0} \int_0^R k r^{ a } \times 4 \pi r^2 d r$

$\text { or } E \times 4 \pi R^2=\left(\frac{4 \pi k}{\in_0}\right) \frac{R^{(a+3)}}{( a +3)}$

$\text { For } r =\frac{ R }{2} \cdot E _2=\frac{ k \left(\frac{R}{2}\right)^{a+1}}{\in_0(a+3)}$

Given, $E _2=\frac{E_1}{8}$

$\text { or } \frac{k\left(\frac{R}{2}\right)^{a+1}}{\in_0(a+3)}=\frac{1 k R^{(a+1)}}{8 \in_0(a+3)}$

$\therefore 2^{\frac{1}{a+1}}=\frac{1}{8}$

or $a =2$.

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