Question 6: Electrostatics & Capacitance

Question

A capacitor of capacitance $$15 \mu \mathrm{F}$$ having dielectric slab of $$\varepsilon_r=2.5$$, dielectric strength $$30 \mathrm{~MV} / \mathrm{m}$$ and potential difference $$=30 \mathrm{~V}$$. Calculate the area of the plate.

Accepted Answer

Given, capacitance of a capacitor
$$\begin{aligned}
& C=15 \mathrm{~nF}=15 	imes 10^{-9} \mathrm{~F} \
& \varepsilon_r=2.5
\end{aligned}$$
Electric field, $$E=30 	imes 10^6 \mathrm{~V} / \mathrm{m}$$
Potential difference, $$V=30 \mathrm{~V}$$
Area of plate $$=$$ ?
If $$d$$ be the distance between the plates, then
$$d=\frac{V}{E}=\frac{30}{30 	imes 10^6}=10^{-6} \mathrm{~m}$$
Capacitance of capacitor,
$$\begin{aligned}
C & =\frac{\varepsilon_0 \varepsilon_r A}{d} \
15 	imes 10^{-9} & =\frac{8.85 	imes 10^{-12} 	imes 2.5 	imes A}{10^{-6}} \
\Rightarrow \quad A & =6.7 	imes 10^{-4} \mathrm{~m}^2
\end{aligned}$$

Answer

$$6.7 	imes 10^{-4} \mathrm{~m}^2$$

Answer

$$4.2 	imes 10^{-4} \mathrm{~m}^2$$

Answer

$$8.0 	imes 10^{-4} \mathrm{~m}^2$$

Answer

$$9.85 	imes 10^{-4} \mathrm{~m}^2$$

Electrostatics & Capacitance

A capacitor of capacitance $15 \mu \mathrm{F} having dielectric slab of \varepsilon_r=2.5, dielectric strength 30 \mathrm{~MV} / \mathrm{m} and potential difference =30 \mathrm{~V}$. Calculate the area of the plate.

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