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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2022
  • Medium
  • MCQ

Two identical thin metal plates has charge \(q _{1}\) and \(q _{2}\) respectively such that \(q _{1}> q _{2}\). The plates were brought close to each other to form a parallel plate capacitor of capacitance \(C\). The potential difference between them is.

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Answer & Solution

Correct Answer

(C) \(\frac{\left(q_{1}-q_{2}\right)}{2 C}\)

Step-by-step Solution

Detailed explanation

Flux \(\phi = EA=\frac{ q _{1}- q _{2}}{2 \varepsilon_{0}}\)

Electric field between plates \(E =\frac{ q _{1}- q _{2}}{2 A \varepsilon_{0}}\)

\(V = Ed =\frac{ q _{1}- q _{2}}{2 A \in_{0}} d\)

\(V =\frac{ q _{1}- q _{2}}{2 C }\)

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