TS EAMCET · Physics · Capacitance
A parallel plate capacitor of capacity \(100 \mu \mathrm{F}\) is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is
- A \(\frac{125}{2} \times 10^{-3}\)
- B \(125 \times 10^{-3}\)
- C \(1.25 \times 10^{-3}\)
- D \(0.0125 \times 10^{-3}\)
Answer & Solution
Correct Answer
(A) \(\frac{125}{2} \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
\[ C=100 \mu \mathrm{F}, V=50 \text { volt } \] Capacitance of parallel plate capacitor \[ C=\frac{\varepsilon_0 A}{d} \] \(d=\) separation between the plates Initial energy…
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