Question 553: Electrostatics & Capacitance

Question

The force between the plates of a parallel plate capacitor of capacitance '$$\mathrm{C}$$' and distance of separation of the plates '$$\mathrm{d}$$' with a potential difference '$$\mathrm{V}$$' between the plates is

Accepted Answer

The force of attraction between the plates of a parallel plate capacitor is
$$\begin{aligned}
& \mathrm{F}=\frac{\sigma^2 \mathrm{~A}}{2 \varepsilon_0}=\frac{\mathrm{Q}^2}{\mathrm{~A}^2} 	imes \frac{\mathrm{A}}{2 \varepsilon_0}\left(\because \sigma=\frac{\mathrm{Q}}{\mathrm{A}}ight) \
& \quad=\frac{\mathrm{Q}^2}{2 \varepsilon_0 \mathrm{~A}}=\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \varepsilon_0 \mathrm{~A}}(\because \mathrm{Q}=\mathrm{CV}) \
& 	herefore \mathrm{F}=\mathrm{C}\left[\frac{\mathrm{CV}^2}{2 \varepsilon_0 \mathrm{~A}}ight]=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} 	imes \frac{\mathrm{CV}^2}{2 \varepsilon_0 \mathrm{~A}} \
& 	herefore \mathrm{F}=\frac{1}{2} \frac{\mathrm{CV}^2}{\mathrm{~d}}
\end{aligned}$$

Answer

$$\frac{\mathrm{V}^2 \mathrm{~d}}{\mathrm{C}}$$

Answer

$$\frac{\mathrm{C}^2 \mathrm{~V}^2}{\mathrm{~d}^2}$$

Answer

$$\frac{\mathrm{CV}^2}{2 \mathrm{~d}}$$

Answer

$$\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \mathrm{~d}^2}$$

Electrostatics & Capacitance

The force between the plates of a parallel plate capacitor of capacitance '$\mathrm{C}' and distance of separation of the plates '\mathrm{d}' with a potential difference '\mathrm{V}$' between the plates is

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