Electric Field and Potential Variation for Uniformly Charged Thin Spherical Shell

To find the variation of the magnitude of the electric field $|\vec{E}(r)|$ and the electric potential $V(r)$ with distance $r$ from the center of a thin spherical shell of radius $R$ carrying a uniform positive charge $Q$:

1. Electric Field $|\vec{E}(r)|$: Applying Gauss's Law, taking a concentric spherical Gaussian surface of radius $r$:

   • For points inside the shell ($r < R$): The enclosed charge is $q_{\text{enclosed}} = 0$. Therefore, by Gauss's Law: $|\vec{E}(r)| = 0$
   • For points outside the shell ($r > R$): The entire charge $Q$ is enclosed: $\oint \vec{E} \cdot d\vec{A} = |\vec{E}(r)|(4\pi r^2) = \frac{Q}{\varepsilon_0} \implies |\vec{E}(r)| = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$
   • Thus, $|\vec{E}(r)|$ is zero for $r < R$, has a jump discontinuity at the surface $r = R$, and falls off as $\frac{1}{r^2}$ for $r > R$. This is represented by the solid curve.

2. Electric Potential $V(r)$: The electric potential is related to the electric field by $V(r) = -\int_\infty^r \vec{E} \cdot d\vec{r}$, with $V(\infty) = 0$:

   • For points outside the shell ($r \ge R$): $V(r) = -\int_\infty^r \frac{1}{4\pi\varepsilon_0}\frac{Q}{r'^2} \, dr' = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$
   • For points inside the shell ($r \le R$): Since $\vec{E} = 0$ inside, the potential remains constant and equal to its value at the surface: $V(r) = V(R) = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R} = \text{constant}$
   • Thus, $V(r)$ is constant for $0 \le r \le R$ and decreases continuously as $\frac{1}{r}$ for $r > R$. This is represented by the dashed curve.

Graph (D) correctly depicts:

• The solid curve being zero for $r < R$ and decaying as $1/r^2$ for $r > R$, representing $|\vec{E}(r)|$.
• The dashed curve being constant for $r \le R$ and decaying as $1/r$ for $r > R$, representing $V(r)$.

Hence, the correct option is D.