Question 533: Electrostatics & Capacitance

Question

Two dielectric slabs having dielectric constant '$$\mathrm{K}_1$$' and '$$\mathrm{K}_2$$' of thickness $$\frac{\mathrm{d}}{4}$$ and $$\frac{3 \mathrm{~d}}{4}$$ are inserted between the plates as shown in figure. The net capacitance between $$A$$ and $$B$$ is $$\left[\varepsilon_0\right.$$ is permittivity of free space]

Accepted Answer

Capacity of $$1^{	ext {st }}$$ Capacitor,
$$C_1=\frac{K_1 \varepsilon_0 A}{d / 4}=\frac{4 K_1 \varepsilon_0 A}{d}$$
Capacity of $$2^{	ext {nd }}$$ Capacitor,
$$\mathrm{C}_2=\frac{\mathrm{K}_2 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d} / 4}=\frac{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d}}$$
Equivalent capacitance $$\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}$$
$$\begin{aligned}
\frac{1}{\mathrm{C}_1} & =\frac{\mathrm{d}}{4 \mathrm{~K}_1 \varepsilon_0 \mathrm{~A}} ; \frac{1}{\mathrm{C}_2}=\frac{3 \mathrm{~d}}{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}} \
\frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \mathrm{~K}_1 \varepsilon_0 \mathrm{~A}}+\frac{3 \mathrm{~d}}{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}} \
\frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \varepsilon_0 \mathrm{~A}}\left[\frac{1}{\mathrm{~K}_1}+\frac{3}{\mathrm{~K}_2}ight] \
\frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \varepsilon_0 \mathrm{~A}}\left[\frac{\mathrm{K}_2+3 \mathrm{~K}_1}{\mathrm{~K}_1 \mathrm{~K}_2}ight] \
	herefore \quad \mathrm{C} & =\frac{4 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left[\frac{\mathrm{K}_1 \mathrm{~K}_2}{3 \mathrm{~K}_1+\mathrm{K}_2}ight]
\end{aligned}$$

Answer

$$\frac{2 \mathrm{~A} \varepsilon_0}{\mathrm{~d}}\left[\frac{\mathrm{K}_1 \mathrm{~K}_2}{3 \mathrm{~K}_1+\mathrm{K}_2}\right]$$

Answer

$$\frac{3 \mathrm{~A} \varepsilon_0}{\mathrm{~d}}\left[\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\right]$$

Answer

$$\frac{3 \mathrm{~A}_0}{2 \mathrm{~d}}\left[\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\right]$$

Answer

$$\frac{4 A \varepsilon_0}{d}\left[\frac{K_1 K_2}{3 K_1+K_2}\right]$$

Electrostatics & Capacitance

Two dielectric slabs having dielectric constant '$\mathrm{K}_1' and '\mathrm{K}_2' of thickness \frac{\mathrm{d}}{4} and \frac{3 \mathrm{~d}}{4} are inserted between the plates as shown in figure. The net capacitance between A and B is \left[\varepsilon_0\right.$ is permittivity of free space]

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