Question

A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is $A{m}^2$ and the separation is $d$ metre. The dielectric constants are $K_1$ and $K_2$ respectively. Its capacitance (in farad) will be :

• A. $\frac{2{\in }_{0}A}{d}\left(\frac{K_1+K_2}{K_1{K}_2}\right)$
• B. $\frac{2{\in }_{0}A}{d}\left(\frac{K_1{K}_2}{K_1+K_2}\right)$
• C. $\frac{{\in }_{0}A}{d}\left(\frac{K_1+K_2}{2K_1{K}_2}\right)$
• D. $\frac{{\in }_{0}A{K}_1{K}_2}{2(d_2{K}_1+d_1{K}_2)}$

Answer 1

The correct option is B $\frac{2{\in }_{0}A}{d}\left(\frac{K_1{K}_2}{K_1+K_2}\right)$

Capacitance of first half of the capacitor of width $d/2=C_1=\frac{K_{1}A{ϵ}_0}{d/2}=\frac{2K_{1}A{ϵ}_0}{d}$

Similarly the capacitance of other half $=C_2=\frac{2K_{2}A{ϵ}_0}{d}$

These two capacitance are connected in series.

Thus, the net capacitance of the combination is, $C_{eff}=\frac{C_1{C}_2}{C_1+C_2}=\frac{2A{ϵ}_0}{d}\left(\frac{K_1{K}_2}{K_1+K_2}\right)$