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A parallel plate capacitor with air as the dielectric has capacitance $C$. A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be
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The correct answer is:
$(K+3) \frac{C}{4}$
The condenser with air as the dielectric has capacitance
$C_1=\frac{\varepsilon_0}{d}\left(\frac{3 A}{4}\right)=\frac{3 \varepsilon_0 A}{4 d}$
Similarly, the condenser with $K$ as the dielectric constant has capacitance
$C_2=\frac{\varepsilon_0 K}{d}\left(\frac{A}{4}\right)=\frac{\varepsilon_0 A K}{4 d}$
Since, $C_1$ and $C_2$ are in parallel
$C_{\text {net }}=C_1+C_2$
$=\frac{3 \varepsilon_0 A}{4 d}+\frac{\varepsilon_0 A K}{4 d}=\frac{\varepsilon_0 A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]$
$=\frac{C}{4}(K+3)$
$C_1=\frac{\varepsilon_0}{d}\left(\frac{3 A}{4}\right)=\frac{3 \varepsilon_0 A}{4 d}$
Similarly, the condenser with $K$ as the dielectric constant has capacitance
$C_2=\frac{\varepsilon_0 K}{d}\left(\frac{A}{4}\right)=\frac{\varepsilon_0 A K}{4 d}$
Since, $C_1$ and $C_2$ are in parallel
$C_{\text {net }}=C_1+C_2$
$=\frac{3 \varepsilon_0 A}{4 d}+\frac{\varepsilon_0 A K}{4 d}=\frac{\varepsilon_0 A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]$
$=\frac{C}{4}(K+3)$
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