Question 92 of 180

A parallel plate capacitor with plate area $A$ and plate separation $d=2m$ has a capacitance of $4\mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K=3$ (as shown in figure) will be

A$2\mu F$B$32\mu F$C$6\mu F$D$8\mu F$

Show answer and solution▾

Answer: (C) $6\mu F$

This parallel plate capacitor can be divided into two capacitors. One with dielectric $\left({C}_{1}\right)$ and other without dielectric$\left({C}_{2}\right)$. The two capacitors will be in series.
Initially $C=\dfrac{{\epsilon }_{0}A}{d}=4\mu F$
Now, ${C}_{1}=\dfrac{k{\epsilon }_{0}A}{\dfrac{d}{2}}=\dfrac{2\times 3\times {\epsilon }_{0}A}{d}=24\mu F$ and ${C}_{2}=\dfrac{{\epsilon }_{0}A}{\dfrac{d}{2}}=\dfrac{2\times {\epsilon }_{0}A}{d}=8\mu F$
Finally ${C}^{'}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}=\dfrac{24\times 8}{24+8}=6\mu F$