A parallel plate capacitor with air as the medium between the plates has a capacitance of $10 \mu \mathrm{F}$. The area of the capacitor is divided into two equal halves and filled with two media having dielectric constant $\mathrm{k} 1=2$ and $\mathrm{k} 2=4$ the capacitance will now be?

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Aicharya · Accepted Answer

Initial capacitance is given as $\mathrm{C}=10 \mu \mathrm{F}$ $\Rightarrow \mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=10 \mu \mathrm{F}$ New arrangement acts as two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ connected in parallel. Thus capacitance of each part $\mathrm{C}_{1}=\frac{\mathrm{k}_{1} \varepsilon_{0} \mathrm{~A}_{1}}{\mathrm{~d}}$ and $\mathrm{C}_{2}=\frac{\mathrm{k}_{2} \varepsilon_{0} \mathrm{~A}_{2}}{\mathrm{~d}}$ Equivalent capacitance of parallel combination $\mathrm{C}_{\mathrm{p}}=\mathrm{C}_{1}+\mathrm{C}_{2}=\left(\mathrm{k}_{1}+\right.$ $\begin{array}{l} \left.\mathrm{k}_{2}\right) \frac{\epsilon_{0}\left(\mathrm{~A}_{1}+\mathrm{A}_{2}\right)}{\mathrm{d}} \ \Rightarrow \mathrm{C}_{\mathrm{p}}=\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right) \frac{\epsilon_{0} \mathrm{~A}}{2 \mathrm{~d}} \quad\left(\because \mathrm{A}_{1}=\mathrm{A}_{2}=\frac{\mathrm{A}}{2}\right) \ \text { Thus } \mathrm{C}_{\mathrm{p}}=(2+4) \times \frac{10}{2}=30 \mu \mathrm{F} \end{array}$

A parallel plate capacitor with air as the medium between the plates has a capacitance of \(10 \mu…

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