Question 543: Electrostatics & Capacitance

Question

The capacitance of a parallel plate capacitor is $$2.5 ~\mu \mathrm{F}$$. When it is half filled with a dielectric as shown in figure, its capacitance becomes $$5 ~\mu \mathrm{F}$$. The dielectric constant of the dielectric is

Accepted Answer

Given $$C=\frac{\varepsilon_0 A}{d}=2.5 ~\mu F$$
When half filled with air,
$$\mathrm{C}_1=\frac{\varepsilon_0(\mathrm{~A} / 2)}{\mathrm{d}}=\frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}} \quad \ldots . .\left(\because \varepsilon_{\mathrm{r}}=1ight)$$
When half filled with a dielectric,
$$\mathrm{C}_2=\frac{\varepsilon_{\mathrm{r}} \varepsilon_0(\mathrm{~A} / 2)}{\mathrm{d}}=\frac{\varepsilon_{\mathrm{r}} \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}$$
From the figure, it can be seen that $$C_1$$ and $$C_2$$ are in parallel configuration.
$$\begin{array}{ll}
	herefore \quad & \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2 \
& 	ext { Given, } \mathrm{C}_{\mathrm{eq}}=5 \mu \mathrm{F} \
& \Rightarrow 5 \mu \mathrm{F}=\frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}+\frac{\varepsilon_{\mathrm{r}} \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}} \
& 5=\frac{2.5}{2}+\varepsilon_{\mathrm{r}} \frac{2.5}{2} \
& \frac{3.75}{1.25}=\varepsilon_{\mathrm{r}} \
	herefore \quad & \varepsilon_{\mathrm{r}}=3
\end{array}$$

Answer

7.5

Answer

3

Answer

4

Answer

5

Electrostatics & Capacitance

The capacitance of a parallel plate capacitor is $2.5 ~\mu \mathrm{F}. When it is half filled with a dielectric as shown in figure, its capacitance becomes 5 ~\mu \mathrm{F}$. The dielectric constant of the dielectric is

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