Question

The figure shows a capacitor having these layers of equal thickness and same areas a that of a plate. Layer 1 is $\mathrm{vacuum}$ . layer II conductor and layer II is a dielectric of dielectric constant K The ratio of energy stored in region III to total energy stored in the capacitor is :

• A. $\frac{1}{K+1}$
• B. $\frac{3}{K+1}$
• C. $\frac{4}{K+3}$
• D. $\frac{4}{3K+1}$

Answer 1

The correct option is A $\frac{1}{K+1}$

$\begin{array}{c}{C}_{eq}=\frac{\frac{E_{0}A}{d}.\frac{k{E}_{0}A}{d}}{\frac{E_{0}B}{d}.\frac{k{E}_{0}A}{d}}=\frac{k\left(\frac{E_{0}A}{d}\right)}{\left(k+1\right)}=\left(\frac{k}{k-1}\right)\left(\frac{E_{0}A}{d}\right)\\ \therefore Q=C_{eq}.v=\left(\frac{k}{k+1}\right)\left(\frac{E_{0}A}{d}\right).v\\ TE=\frac{Q^2}{2C_{eq}}energyinregion=3=\frac{Q^2}{2C}\\ \frac{energyinregion3}{TE}=\frac{C_{eq}}{C}=\frac{\left(\frac{k}{k+1}\right)\left(\frac{E_{0}A}{d}\right)}{k\left(\frac{E_{0}A}{d}\right)}\\ =\frac{1}{k+1}\end{array}$

$\therefore$ Option $A$ is correct .