JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A \(600\,pF\) capacitor is charged by \(200\,V\) supply. It is then disconnected from the supply and is connected to another uncharged \(600\,pF\) capacitor. Electrostatic energy lost in the process is \(.........\,\mu J\).
- A \(6\)
- B \(5\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(6\)
Step-by-step Solution
Detailed explanation
\(Q = CV =600 \times 10^{-12} \times 200=12 \times 10^{-8} C\) \(\text { Initial energy }=\frac{1}{2} CV ^2\) \(=\frac{1}{2} \times 600 \times 10^{-12} \times(200)^2=12\,\mu J\) When connected to another uncharged capacitor Charge will be equally distributed on identical…
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