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A parallel plate capacitor with air between the plates has a capacitance of $8 \mathrm{pF}$. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. (k =6)
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$96 \mathrm{pF}$
Capacity of parallel plate capacitor
$$
\mathrm{C}=\frac{\mathrm{k} \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \quad\left(\text { For air } \mathrm{k}_{\mathrm{r}}=1\right)
$$
So, $\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=8 \times 10^{-12}$
If $d \rightarrow \frac{d}{2}$ and $k_r \rightarrow 6$ then new capacitance $\mathrm{C}^{\prime}=6 \times \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=12 \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=12 \times 8 \mathrm{pF}=96 \mathrm{pF}$
$$
\mathrm{C}=\frac{\mathrm{k} \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \quad\left(\text { For air } \mathrm{k}_{\mathrm{r}}=1\right)
$$
So, $\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=8 \times 10^{-12}$
If $d \rightarrow \frac{d}{2}$ and $k_r \rightarrow 6$ then new capacitance $\mathrm{C}^{\prime}=6 \times \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=12 \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=12 \times 8 \mathrm{pF}=96 \mathrm{pF}$
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