Condition for Linear Radial Dependence of Nuclear Electric Field

According to Gauss's law for a spherically symmetric charge distribution, the radial electric field $E(r)$ at a distance $r$ from the center of the nucleus is related to the enclosed charge $Q_{\text{enclosed}}(r)$ by:

$E(r) = \frac{1}{4\pi\varepsilon_0 r^2} Q_{\text{enclosed}}(r)$

The enclosed charge inside a sphere of radius $r$ is:

$Q_{\text{enclosed}}(r) = \int_0^r 4\pi (r')^2 \rho(r') \, dr'$

Substituting $Q_{\text{enclosed}}(r)$ into the expression for $E(r)$, we get:

$E(r) = \frac{1}{\varepsilon_0 r^2} \int_0^r (r')^2 \rho(r') \, dr'$

For the electric field within the nucleus to be linearly dependent on $r$ (i.e., $E(r) = C r$, where $C$ is a constant), we have:

$C r = \frac{1}{\varepsilon_0 r^2} \int_0^r (r')^2 \rho(r') \, dr'$

$\int_0^r (r')^2 \rho(r') \, dr' = C \varepsilon_0 r^3$

Differentiating both sides with respect to $r$:

$r^2 \rho(r) = 3 C \varepsilon_0 r^2$

$\rho(r) = 3 C \varepsilon_0 = \text{constant}$

This demonstrates that a linear radial dependence of the electric field ($E \propto r$) requires the volume charge density $\rho(r)$ to be uniform throughout the entire volume of the nucleus ($0 \le r \le R$).

From the provided graph:

• $\rho(r) = d$ for $0 \le r \le a$
• $\rho(r)$ decreases linearly from $d$ to $0$ for $a \le r \le R$

For $\rho(r)$ to be constant over the entire region $0 \le r \le R$, the parameter $a$ must be equal to $R$:

$a = R$

Thus, the correct option is C.