Question

The figure shows a parallel plate capacitor with charge $Q_1$ and plate area A. Capacitor is filled with two dielectrics $k_1$ and $k_2$ as shown.

• A. Ratio of energy density in first dielectric to second dielectric is $\frac{5}{3}$
• B. Ratio of energy stored in the two dielectrics is 1 : 1
• C. Surface charge at interface of the two dielectrics is $\frac{2Q}{15}$
• D. Surface charge at interface of the two dielectrics is 0.

Answer 1

Answer: (A), (C)

Surface charge at interface of the two dielectrics is $\frac{2Q}{15}$



$Q_1=Q(1-\frac{1}{k_1})$ and $Q_2=Q(1-\frac{1}{k_2})$
So, $Q_1-Q_2=-\frac{2Q}{15}$

Energy density in capacitor is = $\frac{Energy}{Volume}=\frac{Q^2}{2C\left(Volume\right)}=\frac{Q^2}{2{ϵ}_o{A}^{2}k}$
Hence, $Energy\propto \frac{1}{k}$
which means that the ratio of energy densities is $\frac{5}{3}$.