MHT CET · Physics · Capacitance
The force between the plates of a parallel plate capacitor of capacitance \(C\) and distance of separation of the plates \(d\) with a potential difference \(V\) between the plate is
- A \(\frac{C V^2}{2 d}\)
- B \(\frac{C^2 V^2}{2 d^2}\)
- C \(\frac{C^2 V^2}{d^2}\)
- D \(\frac{V^2 d}{C}\)
Answer & Solution
Correct Answer
(A) \(\frac{C V^2}{2 d}\)
Step-by-step Solution
Detailed explanation
Total electric field between the plates of the capacitor.
\(E=\frac{V}{d}\)
Then, electric field due to only one plate.
\(E_1=\frac{V}{2 d}\)
\(\therefore\) force of one plate on another.
\(F=E_1 \times Q=E_1 \times C V=\frac{V}{2 d} \times C V=\frac{C V^2}{2 d}\)
\(E=\frac{V}{d}\)
Then, electric field due to only one plate.
\(E_1=\frac{V}{2 d}\)
\(\therefore\) force of one plate on another.
\(F=E_1 \times Q=E_1 \times C V=\frac{V}{2 d} \times C V=\frac{C V^2}{2 d}\)
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