Question 549: Electrostatics & Capacitance

Question

A parallel plate air capacitor has a uniform electric field 'E' in the space between the plates. Area of each plate is A and the distance between the plates is '$$\mathrm{d}$$'. The energy stored in the capacitor is $$\left[\varepsilon_0=\right.$$ permittivity of free space)

Accepted Answer

The intensity of the electric field (E) between two plane parallel sheets of equal and opposite charges is given by $$\mathrm{E}=\frac{\sigma}{\varepsilon_0}$$
$$	herefore \sigma=\mathrm{E} \varepsilon_0$$ where $$\sigma=$$ surface density of charge $$=\frac{\mathrm{Q}}{\mathrm{A}}$$
$$	herefore$$ Charge on either plate of the capacitor is $$\mathrm{Q}=\sigma \mathrm{A}=\varepsilon_0 \mathrm{EA}$$ and $$\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$$
$$	herefore$$ The energy stored in the capacitor is
$$\begin{aligned}
& \mathrm{U}=\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}=\frac{\left(\varepsilon_0 \mathrm{EA}ight)^2}{2 \cdot \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}}=\frac{\varepsilon_0^2 \mathrm{~A}^2 \mathrm{E}^2 	imes \mathrm{d}}{2 \varepsilon_0 \mathrm{~A}} \
& 	herefore \mathbf{U}=\frac{\mathbf{1}}{\mathbf{2}} \varepsilon_0 \mathbf{E}^{\mathbf{2}} \mathbf{A d}
\end{aligned}$$

Answer

$$2 \varepsilon_0 \mathrm{EAd}$$

Answer

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

Answer

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

Answer

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

Electrostatics & Capacitance

A parallel plate air capacitor has a uniform electric field 'E' in the space between the plates. Area of each plate is A and the distance between the plates is '$\mathrm{d}'. The energy stored in the capacitor is \left[\varepsilon_0=\right.$ permittivity of free space)

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