Power Law Volume Charge Density in Solid Sphere

To find the value of $a$, we apply Gauss's law for a spherically symmetric charge distribution.

The charge density within the solid sphere of radius $R$ is given by:
$\rho(r) = \kappa r^a$

By Gauss's Law, the electric flux through a concentric spherical Gaussian surface of radius $r$ ($r \le R$) is:
$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$

Due to spherical symmetry, the electric field $\vec{E}$ is directed radially outward and its magnitude is constant over the Gaussian surface:
$E(r) \cdot 4\pi r^2 = \frac{1}{\varepsilon_0} \int_0^r \rho(r') \cdot 4\pi r'^2 \, dr'$

Substitute the charge density $\rho(r') = \kappa r'^a$:
$E(r) \cdot 4\pi r^2 = \frac{4\pi \kappa}{\varepsilon_0} \int_0^r r'^{a+2} \, dr'$
$E(r) \cdot 4\pi r^2 = \frac{4\pi \kappa}{\varepsilon_0} \left[ \frac{r^{a+3}}{a+3} \right]$

Solving for $E(r)$:
$E(r) = \frac{\kappa}{\varepsilon_0(a+3)} r^{a+1}$

We are given that the electric field at $r = \frac{R}{2}$ is $\frac{1}{8}$ times the electric field at $r = R$:
$E\left(\frac{R}{2}\right) = \frac{1}{8} E(R)$

Substituting the expression for $E(r)$:
$\frac{\kappa}{\varepsilon_0(a+3)} \left(\frac{R}{2}\right)^{a+1} = \frac{1}{8} \cdot \frac{\kappa}{\varepsilon_0(a+3)} R^{a+1}$

Cancelling common terms from both sides:
$\left(\frac{1}{2}\right)^{a+1} = \frac{1}{8}$

Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$, we equate the exponents:
$a + 1 = 3$
$a = 2$