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Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distance as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k$ $=$ $2$. The capacitance of the system between $1$ and $3$ $\&$ $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{{{C_1}}}{{{C_2}}}$ is
$1.67$
$1$
$0.6$
$0.71$
Solution
The capacitance of parallel plate capacitor is $C=\frac{A \epsilon_{0}}{d}$
The capacitance of parallel plate capacitor with dielectric is $C^{\prime}=\frac{A k \epsilon_{0}}{d}=k C=2 C$
Here $1$ and $2$ plate and $3$ and $4$ plate will make the capacitor of same capacitance $C$ and $2$ and $3$ will make the capacitance of $C^{\prime}$
Here the capacitance in between $1$ and $3$ is $C_{1}=\frac{C C^{\prime}}{C+C^{\prime}}=\frac{C .2 C}{C+2 C}=2 C / 3$
and the capacitance in between $2$ and $4$ is $C_{2}=\frac{C C^{\prime}}{C+C^{\prime}}=\frac{C .2 C}{C+2 C}=2 C / 3$
$\therefore \frac{C_{1}}{C_{2}}=1$
