Consider a parallel plate capacitor of area A (of each plate) and separation ' $d$ ' between the plates. If $E$ is the electric field and $\varepsilon_0$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is
Solution
<p>$u = \frac{1}{2}\varepsilon_0 E^2$</p>
<p>The energy density of an electric field in free space is given by the expression above. For a parallel plate capacitor with plate area $A$ and separation $d$, the volume between the plates is</p>
<p>$V = Ad.$</p>
<p>Multiplying the energy density by the volume gives the total energy stored:</p>
<p>$$ U = u \times V = \frac{1}{2}\varepsilon_0 E^2 \times Ad = \frac{1}{2}\varepsilon_0 E^2 Ad. $$</p>
<p>Thus, the potential energy stored in the capacitor is</p>
<p>$\frac{1}{2}\varepsilon_0 E^2 Ad.$</p>
<p>This corresponds to the option:</p>
<p>$\text{Option D: } \frac{1}{2}\varepsilon_0 E^2 Ad.$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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