Question

A parallel plate capacitor of area '$A$' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?

• A. $\frac{3K{\epsilon }_{0}A}{4d}$
• B. $\frac{3K{\epsilon }_{0}A}{2d}$
• C. $\frac{(K+1){\epsilon }_{0}A}{2d}$
• D. $\frac{K(K+3){\epsilon }_{0}A}{2(K+1)d}$

Answer 1

Answer: (D)

$c_1=\frac{(A/2){\epsilon }_0}{d/2}=\frac{A{\epsilon }_0}{d},c_2=K\frac{A{\epsilon }_0}{d},c_3=K\frac{A{ϵ}_0}{2d}$
$\therefore c_{\text{eq. }}=\frac{c_1\times c_2}{c_1+c_2}+c_3=\frac{(3+K)KA{\epsilon }_0}{2d(K+1)}$
$(\because C_1$ and $C_2$ are in series and resultant of these two in parallel with $C_3)$.