Question

In the figure a capacitor is filled with dielectrics. The resultant capacitance is:

• A. $\frac{2A{\epsilon }_0}{d}[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}]$
• B. $\frac{A{\epsilon }_0}{d}[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}]$
• C. $\frac{2A{\epsilon }_0}{d}[k_1+k_2+k_3]$
• D. $\frac{A{\epsilon }_0}{d}[\frac{k_3}{2}+\frac{k_1{k}_2}{k_1+k_2}]$

Answer 1

The correct option is D $\frac{A{\epsilon }_0}{d}[\frac{k_3}{2}+\frac{k_1{k}_2}{k_1+k_2}]$

Capacitors $C_{1}and{C}_2$ and in series with $C_3$ in parallel with them

$C_1=\frac{k_{1}A{\epsilon }_0\times 2}{2d}=\frac{k_{1}A{\epsilon }_0}{d}{C}_2=\frac{k_{2}A\times 2{\epsilon }_0}{2d}=\frac{k_{2}A{\epsilon }_0}{d}{C}_3=\frac{k_{3}A{\epsilon }_0}{2d}{C}_{eq}=C_3+\frac{C_1{C}_2}{C_1+C_2}{C}_{eq}=\frac{A{\epsilon }_0}{d}[\frac{k_3}{2}+\frac{k_1{k}_2}{k_1+k_2}]$