Q91 / 125 · BITSAT 2015 · Physics

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho$ $=k r^{a}$, 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 $r=R,$ the value of $a$ is

  • A
    3
  • B
    5
  • Correct answer
    C
    2
  • D
    7

Solution

Using Gauss's law, we have $$ \oint \vec{E} \cdot d \vec{A}=\frac{1}{\epsilon_{0}} \int(\rho d v)=\frac{1}{\epsilon_{0}} \int_{0}^{R} k r^{a} \times 4 \pi r^{2} d r $$ or $E \times 4 \pi R^{2}=\left(\frac{4 \pi k}{\epsilon_{0}}\right) \frac{R^{(a+3)}}{(a+3)}$ $$ \therefore \quad E_{1}=\frac{k R^{(a+1)}}{\epsilon_{0}(a+3)} $$ For $r=\frac{R}{2} \cdot E_{2}=\frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_{0}(a+3)}$ Given, $E_{2}=\frac{E_{1}}{8}$ or $\frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_{0}(a+3)}=\frac{1}{8} \frac{k R^{(a+1)}}{\epsilon_{0}(a+3)}$ $$ \begin{array}{ll} \therefore & \quad \frac{1}{a+1}=\frac{1}{8} \\ \text { or } & a=2 . \end{array} $$

Hence, the correct answer is option: 'C'.

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