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(C) The relationship between the electric field $E$ and the potential difference $dV$ over a distance $dr$ is given by $E = -\frac{dV}{dr}$.In the giv

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(C) The relationship between the electric field $E$ and the potential difference $dV$ over a distance $dr$ is given by $E = -\frac{dV}{dr}$.In the given figure,the potential difference between any two consecutive equipotential lines is constant,i.e.,$dV = 10 \ V$.Therefore,the electric field $E$ is inversely proportional to the distance $dr$ between the equipotential lines $(E \propto \frac{1}{dr})$.At point $B$,the equipotential lines are closest to each other,meaning the distance $dr$ is minimum.Since $dr$ is minimum at $B$,the electric field $E$ is maximum at point $B$.

The figure shows lines of constant potential in a region where an electric field is present. The electric field at point $B$ is............

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