Question

In the following figure, the separation between the plates $2d=7\text{mm}$, plate area $A=56{\text{cm}}^2$ and dielectric constant $k=10$. If the emf of the battery connected is $1\text{kV}$, then the charge on the capacitor is
$($Take ${ϵ}_o=9\times {10}^{-12}\text{F/m})$

• A. $82\text{nC}$
• B. $88\text{nC}$
• C. $86\text{nC}$
• D. $84\text{nC}$

Answer 1

The correct option is D $84\text{nC}$


$C_1$ and $C_2$ are in series.
So, $C_{\text{net 1}}=\frac{C_1{C}_2}{C_1+C_2}$

$\Rightarrow C_{\text{net 1}}=\frac{k{ϵ}_{o}A/2d\times 2k{ϵ}_{o}A/2d}{k{ϵ}_{o}A/2d+2k{ϵ}_{o}A/2d}$

$\Rightarrow C_{\text{net 1}}=k{ϵ}_{o}A/3d$

Further, $C_{\text{net 1}}$ and $C_3$ are in parallel.

$\therefore C_{\text{net}}=C_{\text{net 1}}+C_3$

$\Rightarrow C_{\text{net}}=k{ϵ}_{o}A/3d+k{ϵ}_{o}A/4d$

$\Rightarrow C_{\text{net}}=\frac{7k{ϵ}_{o}A}{12d}$

$\Rightarrow C_{\text{net}}=\frac{7\times 10\times 9\times {10}^{-12}\times 56\times {10}^{-4}}{12\times \left(7/2\right)\times {10}^{-3}}$

$\Rightarrow C_{\text{net}}=8.4\times {10}^{-11}\text{F}$

Now, the charge on the capacitor,

$q=C_{\text{net}}V=8.4\times {10}^{-11}\times {10}^3$

$\Rightarrow q=8.4\times {10}^{-8}\text{C}=84\text{nC}$