If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to
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(b) Electrostatic energy density $\frac{{dU}}{{dV}} = \frac{1}{2}K{\varepsilon _0}{E^2}$
$\frac{{dU}}{{dV}} \propto {E^2}$
$\frac{{dU}}{{dV}} \propto {E^2}$
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