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A solid sphere of radius R has a charge ...

A solid sphere of radius R has a charge Q distributed in its volume with a charge density `rho=kr^a`, where k and a are constants and r is the distance from its centre. If the electric field at `r=(R)/(2)` is `1/8` times that `r=R`, find the value of a.

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To solve the problem, we need to find the value of \( a \) given the electric field at \( r = \frac{R}{2} \) is \( \frac{1}{8} \) times the electric field at \( r = R \). The charge density is given as \( \rho = k r^a \). ### Step-by-Step Solution: 1. **Charge Density and Charge Enclosed**: The charge density is given by \( \rho = k r^a \). To find the total charge \( Q \) enclosed within a sphere of radius \( R \), we will integrate the charge density over the volume of the sphere: \[ Q = \int_0^R \rho \, dV = \int_0^R k r^a \, dV ...
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