Question 351: Electrostats & Capacitance

Question

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

Accepted Answer

$$ u = \frac{1}{2}\varepsilon_0 E^2 $$
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
$$ V = Ad. $$
Multiplying the energy density by the volume gives the total energy stored:
$$ U = u 	imes V = \frac{1}{2}\varepsilon_0 E^2 	imes Ad = \frac{1}{2}\varepsilon_0 E^2 Ad. $$
Thus, the potential energy stored in the capacitor is
$$ \frac{1}{2}\varepsilon_0 E^2 Ad. $$
This corresponds to the option:
$$ 	ext{Option D: } \frac{1}{2}\varepsilon_0 E^2 Ad. $$

Answer

$\varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$

Answer

$\frac{3}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$

Answer

$\frac{1}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$

Answer

$\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$

Electrostats & Capacitance

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

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