MHT CET · Physics · Capacitance
Two identical metal plates are given charges \(\mathrm{q}_1\) and \(\mathrm{q}_2\left(\mathrm{q}_2 < \mathrm{q}_1\right)\) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance 'C', the potential difference 'V' between the plates is
- A \(\frac{q_1-q_2}{c}\)
- B \(\frac{q_1+q_2}{c}\)
- C \(\frac{q_1-q_2}{2 C}\)
- D \(\frac{q_1+q_2}{2 C}\)
Answer & Solution
Correct Answer
(C) \(\frac{q_1-q_2}{2 C}\)
Step-by-step Solution
Detailed explanation
\(Q_{eff} = \frac{q_1 - q_2}{2}\) \(V = \frac{Q_{eff}}{C}\)
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