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A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: (Take 1/(4πε₀) = 9×10⁹ SI units) [R = 3 cm, q = 1 μC]
Inside a charged spherical shell, E = 0 everywhere -> Potential is strictly CONSTANT -> V_C - V_P = 0 V (ZERO)!
Consider a Gaussian spherical surface of radius r < R concentric with the shell. Because all charge q resides on the outer surface of the thin shell, the charge enclosed within the Gaussian sphere is zero. By Gauss’s law:
The electric field is the negative gradient of the electrostatic potential:
Because potential is constant everywhere from the center up to the surface, the potential at the center C is identically equal to the potential at point P on the surface:
The potential difference between points C and P is therefore identically zero:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Arbitrary non-zero value.
Fundamental electrostatic property: hollow shell interior is an equipotential volume.
This is the absolute potential V_P = 9×10⁹ × 10⁻⁶ / 0.03 = 3 × 10⁵ V, NOT the difference!
Incorrect non-zero value.
Candidates calculate V = kq/R = 3 × 10⁵ V and choose Option (3), forgetting that the question specifically asks for the potential DIFFERENCE between C and P!
"The entire interior of a uniformly charged thin spherical shell is equipotential, with potential everywhere equal to that on its surface."
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