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Physics-Potential Difference

NEET Physics Potential Difference MCQ Question

Type: MCQ-numerical-Medium-Class 12

Given a parallel plate capacitor with a dielectric fully inserted between its plates, if the surface charge density on the plates is σ\sigma and the dielectric reduces the electric field by a factor of 2, what is the new potential difference across the plates?

A

σd2ε0\frac{\sigma d}{2\varepsilon_0}

B

2σdε0\frac{2\sigma d}{\varepsilon_0}

C

σdε0\frac{\sigma d}{\varepsilon_0}

D

σd4ε0\frac{\sigma d}{4\varepsilon_0}

Correct Answer

Option A

Detailed Explanation

With a dielectric, the potential difference is V=(σ−σp)dε0V = \frac{(\sigma - \sigma_p) d}{\varepsilon_0}, and if the field is reduced by a factor of 2, σ−σp=σ2\sigma - \sigma_p = \frac{\sigma}{2}. Thus, V=σd2ε0V = \frac{\sigma d}{2\varepsilon_0}.

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