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

$ρ = K r^{a}; E (r = \frac{R}{2}) = \frac{1}{8} E (r = R) \frac{q_{e n c l o s e d}}{4 π ε_{0} {(R / 2)}^{2}} = \frac{1}{8} \frac{Q}{4 π ε_{0} R^{2}} 32 q_{e n c l o s e d} = Q q_{e n c l o s e d} = \int_{0}^{R / 2} k r^{a} 4 π r^{2} d r = \frac{4 π k}{(a + 3)} {(\frac{R}{2})}^{(a + 3)}, Q = \frac{4 π k}{(a + 3)} R^{(a + 3)}, \frac{Q}{q_{e n c l o s e d}} = 2^{a + 3} 2^{a + 3} = 32 \Rightarrow a = 2$