The variation of electrostatic potential with radial distance $r$ from the centre of a positively charged meta

Question

The variation of electrostatic potential with radial distance $r$ from the centre of a positively charged metallic thin shell of radius $R$ is given by the graph:

Accepted Answer

1. **Inside the Shell ($r < R$):** According to the property of **electrostatic shielding** (mentioned in Class 11 Physics, Points to Ponder [1]), the electric field $E$ inside a charged hollow conductor is zero. Since the electric field is the gradient of the potential ($E = -dV/dr$), a zero field implies that the potential $V$ is constant throughout the interior. Its value is equal to the potential at the surface: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$.
2. **Outside the Shell ($r \geq R$):** For points outside, the shell behaves as if the entire charge $Q$ were concentrated at its center (analogous to the gravitational case for a spherical shell discussed in Class 11 Physics [2]). Thus, the potential decreases inversely with distance: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$.
3. **Graph Characteristics:** The correct graph starts at a positive constant value for $r$ from $0$ to $R$, and then decreases as a hyperbola ($1/r$) for $r > R$. This corresponds to the standard behavior described in Option 2.

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