Question

Three dielectrics of thickness $t_1,t_2,t_3$ and dielectric constants $k_1,k_2,k_3$ respectively are placed in between the plates of a
capacitor as shown. What is the new capacitance?

• A. $\frac{{\epsilon }_{0}A}{d}$
• B. $\frac{{\epsilon }_0{k}_1{k}_2{k}_{3}A}{d-(t_1+t_2+t_3)}$
• C. $\frac{{\epsilon }_{0}A}{d-(t_1+t_2+t_3)+\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}}$
• D. None of these

Answer 1

The correct option is C $\frac{{\epsilon }_{0}A}{d-(t_1+t_2+t_3)+\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}}$

We know that on insertion of a dielectric of thickness d within a capacitor of plate separation d, the capacitance increases by $C=\frac{K{\epsilon }_{0}A}{d}$ This is actually a common form of a general formula where if you have dielectrics of different thicknesses as given in the question the capacitance becomes $\frac{{\epsilon }_{0}A}{d-(t_1+t_2+t_3)+\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}}$. From where the hell this formula comes? If you have that question, I recommend you to watch the derivation video once again, it must have skipped your attention!