Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

Option 1

Option 2

Option 3

Option 4

Step-by-Step Solution

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: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$ .
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: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$ .
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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