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1. Electric Charges and Fields
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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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