Analysis of Electric Field $|"`E`(r)|$ and Electric Potential V(r) for a Charged Spherical Shell

We are given a thin spherical shell of radius $R$ with a uniform positive surface charge density $\sigma$. Let the total charge on the shell be $Q = \sigma \cdot (4\pi R^2)$. We need to find the variation of the electric field magnitude $|"`E`(r)|$ and electric potential $V(r)$ with distance $r$ from the center.

1. Electric Field $|"`E`(r)|$

We use Gauss's Law, $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$, to determine the electric field.

Case 1: Inside the shell ($r < R$)

• We construct a spherical Gaussian surface of radius $r < R$ concentric with the shell.
• The charge enclosed by this surface, $q_{enc}$, is zero, as all the charge resides on the surface of the shell at radius $R$.
• From Gauss's Law, $\oint \vec{E} \cdot d\vec{A} = 0$. Due to spherical symmetry, $E \cdot (4\pi r^2) = 0$.
• This implies that the electric field inside the shell is zero: $E(r) = 0$ for $r < R$.

Case 2: Outside the shell ($r > R$)

• We construct a spherical Gaussian surface of radius $r > R$.
• The charge enclosed, $q_{enc}$, is the total charge of the shell, $Q$.
• Applying Gauss's Law: $E \cdot (4\pi r^2) = \frac{Q}{\epsilon_0}$.
• Solving for $E$, we get $E(r) = \frac{Q}{4\pi \epsilon_0 r^2}$ for $r > R$. The electric field outside the shell is the same as that of a point charge $Q$ located at the center.

At the surface ($r = R$)

• Just inside the surface ($r \to R^-$), $E = 0$.
• Just outside the surface ($r \to R^+$), $E = \frac{Q}{4\pi \epsilon_0 R^2}$.
• There is a discontinuity in the electric field at $r = R$.

Summary for $|"`E`(r)|$:

• $|"`E`(r)| = 0$ for $r < R$
• $|"`E`(r)|$ jumps to $\frac{Q}{4\pi \epsilon_0 R^2}$ at $r = R$
• $|"`E`(r)| = \frac{Q}{4\pi \epsilon_0 r^2}$ (proportional to $1/r^2$) for $r > R$

2. Electric Potential $V(r)$

The electric potential is related to the electric field by $V = -\int \vec{E} \cdot d\vec{r}$. We take the reference potential $V(\infty) = 0$.

Case 1: Outside the shell ($r \ge R$)

• $V(r) - V(\infty) = -\int_{\infty}^{r} E(r') dr'$
• $V(r) = -\int_{\infty}^{r} \frac{Q}{4\pi \epsilon_0 r'^2} dr' = -\frac{Q}{4\pi \epsilon_0} \left[ -\frac{1}{r'} \right]_{\infty}^{r} = \frac{Q}{4\pi \epsilon_0 r}$
• So, $V(r) = \frac{Q}{4\pi \epsilon_0 r}$ for $r \ge R$.

Case 2: Inside the shell ($r \le R$)

• We know that inside the shell, $E = -\frac{dV}{dr} = 0$. This implies that the potential $V$ is constant for $r < R$.
• Since potential is a continuous function, the potential inside must be equal to the potential at the surface ($r=R$).
• The potential at the surface is $V(R) = \frac{Q}{4\pi \epsilon_0 R}$.
• Therefore, $V(r) = \frac{Q}{4\pi \epsilon_0 R}$ (constant) for $r \le R$.

Summary for $V(r)$:

• $V(r) = \frac{Q}{4\pi \epsilon_0 R}$ (constant) for $r \le R$
• $V(r) = \frac{Q}{4\pi \epsilon_0 r}$ (proportional to $1/r$) for $r > R$

3. Matching with Graphs

We look for a graph that represents these behaviors:

• For $|"`E`(r)|$: Zero until $r=R$, then a sharp jump, followed by a $1/r^2$ decay.
• For $V(r)$: A constant non-zero value until $r=R$, then a $1/r$ decay. The function must be continuous at $r=R$.

Let's evaluate the options:

• Option A: Incorrect. Shows non-zero E inside and increasing V inside.
• Option B: Incorrect. Shows linearly increasing E inside, which is characteristic of a solid uniformly charged non-conducting sphere, not a shell.
• Option C: Incorrect. Shows correct E field, but V is zero inside, which is wrong. Potential is constant and non-zero inside.
• Option D: Correct.
  • The graph for $|"`E`(r)|$ is zero for $r<R$, jumps at $r=R$, and decays as $1/r^2$ for $r>R$.
  • The graph for $V(r)$ is constant for $r \le R$ and decays as $1/r$ for $r > R$. The potential graph is continuous at $r=R$.

Thus, the graph in option D correctly represents the variation of both the electric field magnitude and the electric potential.