Question 539: Electrostatics & Capacitance

Question

The mean electrical energy density between plates of a charged air capacitor is (where $$\mathrm{q}=$$ charge on capacitor, $$\mathrm{A}=$$ Area of capacitor plate)

Accepted Answer

For a parallel plate capacitor, the energy density $$=\frac{1}{2} E^2 \varepsilon_0$$
But $$\mathrm{E}=\frac{\sigma}{\varepsilon_0}$$
$$\begin{aligned}
	herefore \quad 	ext { Energy density } & =\frac{1}{2} \frac{\sigma^2}{\varepsilon_0^2} 	imes \varepsilon_0 \
& =\frac{\sigma^2}{2 \varepsilon_0} \
& =\frac{\left(\frac{\mathrm{q}}{\mathrm{A}}ight)^2}{2 \varepsilon_0} \quad \ldots .(\because \sigma=\mathrm{q} / \mathrm{A}) \
& =\frac{\mathrm{q}^2}{2 \mathrm{~A}^2 \cdot \varepsilon_0}
\end{aligned}$$

Answer

$$\frac{q^2}{2 \varepsilon_0 A^2}$$

Answer

$$\frac{\mathrm{q}}{2 \varepsilon_0 \mathrm{~A}^2}$$

Answer

$$\frac{\mathrm{q}^2}{2 \varepsilon_0 \mathrm{~A}}$$

Answer

$$\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{q}^2}$$

Electrostatics & Capacitance

The mean electrical energy density between plates of a charged air capacitor is (where $\mathrm{q}= charge on capacitor, \mathrm{A}=$ Area of capacitor plate)

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