A solid sphere of radius R has a charge Q distributed in its volume with a charge density $ρ = {kr}^{a},$ 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.

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answer is 2.

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Detailed Solution

$\begin{array}{l} ρ = {Kr}^{a}; E (r = \frac{R}{2}) = \frac{1}{8} E (r = R) \\ \frac{q_{enclosed}}{4 {πε}_{0} (R / 2)^{2}} = \frac{1}{8} \frac{Q}{4 {πε}_{0} R^{2}} \Rightarrow 32 q_{enclosed} = Q \\ q_{enclosed} = \int_{0}^{R / 2} {kr}^{a} 4 {πr}^{2} dr = \frac{4 πk}{(a + 3)} {(\frac{R}{2})}^{(a + 3)} \\ Q = \frac{4 πk}{(a + 3)} R^{(a + 3)}, \frac{Q}{q_{enclosed}} = 2^{a + 3} \\ 2^{a + 3} = 32 \Rightarrow a = 2 \end{array}$