AP EAMCET · PHYSICS · Capacitance
A parallel plate capacitor of capacitance \(500 \mathrm{pF}\) charged with \(100 \mathrm{~V}\) supply: It is then disconnected from the supply and connected to another uncharged \(500 \mathrm{pF}\) capacitor. The electrostatic energy lost in this process is
- A \(1.25 \mu \mathrm{J}\)
- B \(0.175 \mu \mathrm{J}\)
- C \(0.225 \mu \mathrm{J}\)
- D \(0.275 \mu \mathrm{J}\)
Answer & Solution
Correct Answer
(A) \(1.25 \mu \mathrm{J}\)
Step-by-step Solution
Detailed explanation
Initial stored energy = Energy of capacitor 1 \(=\frac{1}{2} C_1 V_1^2\)…
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