MHT CET · Physics · Capacitance
The plates of a parallel plate capacitor of capacity ' \(\mathrm{C}_1\) ' are moved closer together until they are half their original separation. The new capacitance ' \(\mathrm{C}_2\) ' is
- A \(\mathrm{C}_2=\frac{\mathrm{C}_1}{2}\)
- B \(\mathrm{C}_2=\mathrm{C}_1\)
- C \(\mathrm{C}_2=2 \mathrm{C}_1\)
- D \(C_2=4 C_1\)
Answer & Solution
Correct Answer
(C) \(\mathrm{C}_2=2 \mathrm{C}_1\)
Step-by-step Solution
Detailed explanation
The capacitance of a parallel plate capacitor is given by
\(
\mathrm{C}=\frac{\mathrm{kA} \varepsilon_0}{\mathrm{~d}}
\)
If the distance between the plates \(\mathrm{d}\) is decreased to half, the capacitance will become double.
\(
\mathrm{C}=\frac{\mathrm{kA} \varepsilon_0}{\mathrm{~d}}
\)
If the distance between the plates \(\mathrm{d}\) is decreased to half, the capacitance will become double.
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