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  • A parallel plate capacitor of plate area   and plate separation <img alt="Y" src="https://entrancecorner.oncodecogs.

A parallel plate capacitor of plate area  X and plate separation Y is charged by a battery which has the voltage V. After this, the battery gets disconnected. Determine the work that is required to pull the plates at a separation of 2 \mathrm{Y} now.

Option: 1

W=\frac{\epsilon_o X V^2}{2 Y}


Option: 2

W=\frac{\epsilon_o Y V^2}{2 X}


Option: 3

W=\frac{\epsilon_o X V}{2 Y}


Option: 4

None of the above.


Answers (1)

best_answer

It is already known that C=\frac{\epsilon_o A}{D}

When separation  D is doubled, new capacitance would be =

\begin{aligned} & C^{\prime}=\frac{\epsilon_o A}{2 D} \\ & C^{\prime}=\frac{C}{2} \end{aligned}

Initial energy, for D = 

U_i=\frac{Q^2}{2 C}

Final energy, for 2 \mathrm{D}=

U_f=\frac{Q^2}{C}

Work = 

=U_f-U_i

=\frac{Q^2}{2 C}

Putting  \mathrm{Q}=\mathrm{CV} \text {; }

\frac{1}{2} C V^2

Putting \mathrm{A}=\mathrm{X} and  \mathrm{D}=\mathrm{Y}according to mentioned in question; 

W=\frac{\epsilon_o X V^2}{2 Y}

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mansi

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