A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ___________ nJ.
Answer (integer)
125
Solution
<p>Electrical energy lost $= {1 \over 2}\left( {{1 \over 2}C{V^2}} \right)$</p>
<p>$= {1 \over 2} \times {1 \over 2} \times 50 \times {10^{ - 12}} \times {(100)^2}$</p>
<p>$= {{500} \over 4}$ nJ</p>
<p>$= 125$ nJ</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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