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A capacitor of capacitance \mathrm{50pF} is charged by \mathrm{100V} source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _______\mathrm{nJ}.

Option: 1

125


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

 

Initial :

Finally

\mathrm{Initial\: energy =\frac{1}{2} C V^{2}=U_{1}}

\mathrm{\text { Final energy } =\frac{1}{2} C\left(\frac{V}{2}\right)^{2}+\frac{1}{2} \times C \times\left(\frac{V}{2}\right)^{2}} \\

                           \mathrm{=\frac{C V^{2}}{4}=U_{2}}

\mathrm{\text { Loss in energy }=U_{1}-U_{2}} \\

                              \mathrm{=\frac{1}{2} C V^{2}-\frac{1}{4} CV^{2}=\frac{C V^{2}}{4}} \\

                              \mathrm{=\frac{1}{4} \times 50 \times 10^{-12} \times 100 \times 100} \\

                              \mathrm{=\frac{50 \times 10^{4} \times 10^{-3} \times 10^{-9}}{4} }\\

                              \mathrm{=125 \mathrm{~nJ} }

Hence answer is \mathrm{125 \mathrm{~nJ} }.

Posted by

Nehul

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