Gujarati
Hindi
1. Electric Charges and Fields
normal

If the distance between the plates of a capacitor having capacity $C$ and charge $Q$ is doubled then work done will be

A

$\frac{{{Q^2}}}{{4C}}$

B

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

C

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

D

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

Solution

$\mathrm{W}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=\frac{\mathrm{Q}^{2}}{2\left(\frac{\mathrm{C}}{2}\right)}-\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$

Standard 12
Physics

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