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

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answer is 2.

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Detailed Solution

Charge enclosed till point P is given by $Q = \int ρ d V = \int_{0}^{R} (k r^{2}) (4 π r^{2} d r) = \frac{4 π k}{a + 3} (R^{a + 3})$

$Q = \int p d V = \int_{0}^{\frac{R}{2}} (k r^{2}) (4 π r^{2} d r) = \frac{4 π k}{a + 3} {(\frac{R}{2})}^{a + 3}$

According to question

$\frac{1}{4 π ε_{0}} \frac{Q}{{(\frac{R}{2})}^{2}} = \frac{1}{8} (\frac{1}{4 π ε_{0}} \frac{Q}{R^{2}})$

Putting the value of Q and Q’ get

a = 2

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