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A capacitor having capacitance $1 \mu \mathrm{F}$ with air, is filled with two dielectrics as shown. How many times capacitance will increase?
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Verified Answer
The correct answer is:
$6$
Key Idea : After filling with dielectrics the two capacitors will be in parallel order.
As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor
$C=\frac{\varepsilon_0 A}{d}$
where $A$ is area of each plate and $d$ is the separation between the plates.
After filling with dielectrics, we have two capacitors of capacitance.
$C_1=\frac{K_1 \varepsilon_0(A / 2)}{d}=\frac{8}{2} \frac{\varepsilon_0 A}{d}=\frac{4 \varepsilon_0 A}{d}=4 C$
and $C_2=\frac{K_2 \varepsilon_0(A / 2)}{d}=\frac{4}{2} \frac{\varepsilon_0 A}{d}=\frac{2 \varepsilon_0 A}{d}=2 C$
Hence, their equivalent capacitance
$\begin{aligned} C_{\mathrm{eq}} & =C_1+C_2 \\ & =4 C+2 C=6 C\end{aligned}$
ie, new capacitance will be six times of the original.
As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor
$C=\frac{\varepsilon_0 A}{d}$
where $A$ is area of each plate and $d$ is the separation between the plates.
After filling with dielectrics, we have two capacitors of capacitance.
$C_1=\frac{K_1 \varepsilon_0(A / 2)}{d}=\frac{8}{2} \frac{\varepsilon_0 A}{d}=\frac{4 \varepsilon_0 A}{d}=4 C$
and $C_2=\frac{K_2 \varepsilon_0(A / 2)}{d}=\frac{4}{2} \frac{\varepsilon_0 A}{d}=\frac{2 \varepsilon_0 A}{d}=2 C$
Hence, their equivalent capacitance
$\begin{aligned} C_{\mathrm{eq}} & =C_1+C_2 \\ & =4 C+2 C=6 C\end{aligned}$
ie, new capacitance will be six times of the original.
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