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A parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu \mathrm{F}$. The area of capacitor is divided into two equal halves and filled with two media (as shown in figure) having dielectric constant $K_1=2$ and $\mathrm{K}_2=4$. The capacitance of the system will be
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The correct answer is:
$30 \mu \mathrm{F}$
$\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=10 \mu \mathrm{F}$
After dividing the area into two equal halves, the resultant capacitance is calculated as,
$\begin{aligned}
\mathrm{C}_{\text {eq }} & =\mathrm{C}_1+\mathrm{C}_2 \\
& =\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}+\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}=\frac{\frac{2 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}}+\frac{\frac{4 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}} \\
\therefore \quad \mathrm{C}_{\text {eq }} & =\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}+\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{3 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=3 \times 10=30 \mu \mathrm{F}
\end{aligned}$
After dividing the area into two equal halves, the resultant capacitance is calculated as,
$\begin{aligned}
\mathrm{C}_{\text {eq }} & =\mathrm{C}_1+\mathrm{C}_2 \\
& =\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}+\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}=\frac{\frac{2 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}}+\frac{\frac{4 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}} \\
\therefore \quad \mathrm{C}_{\text {eq }} & =\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}+\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{3 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=3 \times 10=30 \mu \mathrm{F}
\end{aligned}$
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