JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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.
- A \(\frac{\left(q_{1}+q_{2}\right)}{C}\)
- B \(\frac{\left( q _{1}- q _{2}\right)}{ C }\)
- C \(\frac{\left(q_{1}-q_{2}\right)}{2 C}\)
- D \(\frac{2\left(q_{1}-q_{2}\right)}{C}\)
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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