Two identical parallel plate capacitors, of capacitance C each, have plates of area A separated by a distance d. The space between the plates of two capacitors is filled with dielectrics of equal thickness and dielectric constants \( K_1=1,\ K_2=2 \) and \( K_3=3 \) as shown. If these two modified capacitors are charged by same potential, the ratio of energy stored \( E_1 \), (capacitor I) to \( E_2 \) (capacitor II) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{E_1}{E_2}=\frac{11}{9} \)
The energy stored at a given potential V is \( \frac{1}{2}CV^{2} \), so for the same V the ratio of the energies is the ratio of the capacitances. Each modified capacitor is found from the combination rules, with \( C = \varepsilon_0A/d \).
In the figure, capacitor I has the three dielectrics side by side and capacitor II has them as three layers one above the other.
Capacitor II: three layers one above the other (in series), each of thickness \( d/3 \):
\( \displaystyle \frac{1}{C_s} = \frac{d/3}{\varepsilon_0A}\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3}\right) = \frac{d}{3\varepsilon_0A}\times\frac{11}{6} = \frac{11}{18C} \)
\( \displaystyle C_s = \frac{18}{11}\,C \)
Capacitor I: three sections side by side (in parallel), each of area \( A/3 \):
\( \displaystyle C_p = \frac{\varepsilon_0(A/3)}{d}\,(1 + 2 + 3) = 2C \)
\( \displaystyle \frac{E_1}{E_2} = \frac{C_p}{C_s} = \frac{2C}{18C/11} = \frac{11}{9} \)