Question

A parallel-plate capacitor of capacitance $5\mu F$ is connected to a battery of emf $6V.$ The separation between the plates is $2mm$. A dielectric slab of thickness $1mm$ and dielectric constant $5$ is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination.

Answer 1

$C=5\mu fV=6Vd=2mm=2\times {10}^{-3}m$

$d=2\times {10}^{-3}m$

$t=1\times {10}^{-3}m$

$k=5$ or

$C=\frac{{\epsilon }_{0}A}{d}$

$\Rightarrow 5\times {10}^{-6}=\frac{8.85\times A\times {10}^{-12}}{2\times {10}^{-3}}\times {10}^{-9}\Rightarrow A=\frac{{10}^4}{8.85}$

When the dielectric placed on it

$C_1=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{k}}=\frac{8.85\times {10}^{-12}\times \frac{{10}^4}{8.85}}{2\times {10}^{-3}-{10}^{-3}+\frac{{10}^{-3}}{5}}=\frac{{10}^{-12}\times {10}^4\times 5}{6\times {10}^{-3}}=\frac{5}{6}\times {10}^{-5}=0.00000833=8.33\mu F$