JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 24)
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.
Odpověď
125
Vysvětlení
Electrical energy lost $$ = {1 \over 2}\left( {{1 \over 2}C{V^2}} \right)$$
$$ = {1 \over 2} \times {1 \over 2} \times 50 \times {10^{ - 12}} \times {(100)^2}$$
$$ = {{500} \over 4}$$ nJ
$$ = 125$$ nJ