Question

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. $8\mu F$

Answer 1

Answer: (C)

$C_1=\frac{A{\epsilon }_0}{d/2}=\frac{2A{\epsilon }_0}{d}=C$
$C_2=\frac{KA{\epsilon }_0}{d/2}=\frac{2KA{\epsilon }_0}{d}=\frac{6A{\epsilon }_0}{d}=3C$
$C_1\&C_2$ are in series
$C_{\text{new }}=\frac{C_1{C}_2}{C_1+C_2}=\frac{C\times 3C}{C+3C}=\frac{3C}{4}$
$=\frac{3}{4}\times \frac{2A{\epsilon }_0}{d}=\frac{3}{2}\times \frac{A{\epsilon }_0}{d}$
$C_{\text{new }}=\frac{3}{2}{C}_{\text{original }}$
$=\frac{3}{2}\times 4=6\mu F$