Question

I have multiple dielectrics (of surface area in contact with capacitor as $A_1,A_{2}and{A}_3)$ in my capacitor inserted as shown. What is my new capacitance?

• A. $\frac{{\epsilon }_0}{d}(k_1+k_2+k_3)A$
• B. $\frac{{\epsilon }_{0}d}{A}(k_1+k_2+k_3)$
• C. $\frac{{\epsilon }_0}{d}(k_1{A}_1+k_2{A}_2+k_3{A}_3)$
• D. $\frac{{\epsilon }_0}{d}(A_1+A_2+A_3)k_1$

Answer 1

The correct option is C $\frac{{\epsilon }_0}{d}(k_1{A}_1+k_2{A}_2+k_3{A}_3)$

As all the dielectric share the same conductor on either side, we can consider the situation to be three capacitor in parallel with each other. Something like this

The capacitance of any of these capacitor would be $C=\frac{k{\epsilon }_{0}A}{d}$ So, $Cnet=C_1+C_2+C_3$ (Because all of them are in parallel)
$C_{net}=\frac{{\epsilon }_0}{d}(k_1{A}_1+k_2{A}_2+k_3{A}_3)$