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(A) Initially,the two capacitors of capacitance $C$ are in series. The equivalent capacitance is $C_{eq,i} = \frac{C}{2}$. The initial charge on each

Vedclass · Accepted Answer

$\frac{CV^2}{12}$

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(A) Initially,the two capacitors of capacitance $C$ are in series. The equivalent capacitance is $C_{eq,i} = \frac{C}{2}$. The initial charge on each capacitor is $q_i = C_{eq,i} V = \frac{CV}{2}$. The initial energy stored is $U_i = 2 	imes \frac{1}{2} C (V/2)^2 = \frac{CV^2}{4}$.After inserting the dielectric $(K=2)$ into one capacitor,its new capacitance becomes $C' = KC = 2C$. The new equivalent capacitance is $C_{eq,f} = \frac{C 	imes 2C}{C + 2C} = \frac{2C}{3}$. The new charge on the circuit is $q_f = C_{eq,f} V = \frac{2CV}{3}$.The charge supplied by the battery is $\Delta q = q_f - q_i = \frac{2CV}{3} - \frac{CV}{2} = \frac{CV}{6}$.The work done by the battery is $W_b = \Delta q V = \frac{CV^2}{6}$.The final energy stored is $U_f = \frac{1}{2} C (V_1)^2 + \frac{1}{2} (2C) (V_2)^2$. Since the capacitors are in series,$V_1 + V_2 = V$ and $q_f = C V_1 = 2C V_2$,so $V_1 = 2V_2$. Thus $3V_2 = V$,$V_2 = V/3$ and $V_1 = 2V/3$. $U_f = \frac{1}{2} C (2V/3)^2 + \frac{1}{2} (2C) (V/3)^2 = \frac{1}{2} C (4V^2/9) + C (V^2/9) = \frac{2CV^2}{9} + \frac{CV^2}{9} = \frac{CV^2}{3}$.Using the energy conservation principle: $W_b = \Delta U + H$,where $H$ is the heat loss.$H = W_b - (U_f - U_i) = \frac{CV^2}{6} - (\frac{CV^2}{3} - \frac{CV^2}{4}) = \frac{CV^2}{6} - \frac{CV^2}{12} = \frac{CV^2}{12}$.

Initially,the circuit is in a steady state. Now,one of the capacitors is filled with a dielectric of dielectric constant $K = 2$. Find the heat loss in the circuit due to the insertion of the dielectric.

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