The amount of work done in increasing the voltage across the plates of a capacitor form 5 V to 10 V is ' W '. The work done in increasing it from 10 V to 15 V will be (nearly)

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Aicharya · Accepted Answer

$\mathrm{W}=\frac{1}{2} \mathrm{CV}^2$
For constant $\mathrm{C}, \mathrm{W} \propto \mathrm{V}^2$ $\begin{array}{ll} \therefore & \frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{\mathrm{V}_2^2-\mathrm{V}_1^2}{\mathrm{~V}_3^2-\mathrm{V}_2^2} \\ & \text { here } \mathrm{V}_1=5 \mathrm{~V}, \mathrm{~V}_2=10 \mathrm{~V}, \mathrm{~V}_3=15 \mathrm{~V} \\ \therefore \quad & \frac{\mathrm{~W}}{\mathrm{~W}_2}=\frac{100-25}{225-100}=\frac{75}{125} \\ \therefore \quad & \mathrm{~W}_2=1.67 \mathrm{~W} \end{array}$

The amount of work done in increasing the voltage across the plates of a capacitor form 5 V to 10 V is ' W '…

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