Question

If an additional small charge \(dq\) is supplied to a capacitor when its potential difference is \(V\), what is the small work done?

Options

• \[dW=V\,dq=\frac{q}{C}\,dq\]

• \[dW=\frac{C}{q}\,dq\]

• \[dW=\frac{q}{C}\]

• \[dW=\frac{1}{2}V\,dq\]

Answer 1

The small work done in supplying charge \(dq\) is the potential difference multiplied by that charge increment: \(dW=V\,dq\). Using \(V=\frac{q}{C}\) gives \(dW=\frac{q}{C}\,dq\).