QJEEM22Phb3fb: JEE Main 2022 Physics - Class 12 JEEM22Phb3fb Solution: A capacitor of capacitance 50 pF is charged by 100 V source. It …

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QJEEM22Phb3fb: JEE Main 2022 Physics - Class 12 JEEM22Phb3fb Solution: A capacitor of capacitance 50 pF is charged by 100 V source. It …

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.

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Accepted Answer

Electrical energy lost $= \frac{1}{2} (\frac{1}{2} C V^{2})$

$= \frac{1}{2} \times \frac{1}{2} \times 50 \times 10^{- 12} \times (100)^{2}$

$= \frac{500}{4}$ nJ

$= 125$ nJ

QJEEM22Phb3fb: JEE Main 2022 Physics - Class 12 JEEM22Phb3fb Solution: A capacitor of capacitance 50 pF is charged by 100 V source. It …

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