Change in electrostatics potential energy of the capacitor will be | TestHub

Passage / Comprehension

In the figure conducting plates form a capacitor of initial capacity C and charges on the plates are + Q and -Q respectively. 1 mol monoatomic ideal gas is filled up at initial temperature $T_{0}$ . Di-electric constant of the gas is given by a hypothetical relation $K = K_{0} [1 + (\frac{T}{T _{0}})^{3}]$ where T is absolute temperature of the gas. Now piston is moved slowly by an internal agent so that separation between the plates becomes double the initial separation. Answer the following questions.\\

Change in electrostatics potential energy of the capacitor will be

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Answer:

Solution:

From the adiabatic relation $T V^{γ - 1} = co n s t an t$ :

$\frac{T}{T _{0}} ∴ ini t ia ll y K_{i} an df ina ll y K_{f} ∴ f ina l c a p a c i t y C^{'} ini t ia ll y C ∴ \frac{C ^{'}}{C} ∴ C^{'} ∴ Δ U_{e l ec t ro .} = (\frac{V _{0}}{V})^{γ - 1} = (\frac{1}{2})^{\frac{2}{3}} = K_{0} [1 + (\frac{T _{0}}{T _{0}})^{3}] = 2 K_{0} = K_{0} 1 + {(\frac{1}{2})^{2/3}}^{3} = \frac{5 K _{0}}{4} = \frac{ε _{0} K _{f} A}{2 d} (d = ini t ia l se p a r a t i o n) = \frac{ε _{0} K _{i} A}{d} = \frac{K _{f}}{2 K _{i}} = \frac{5 K _{0} /4}{2 ( 2 K _{0} )} = \frac{5}{16} = \frac{5 C}{16} = \frac{Q ^{2}}{2 C ^{'}} - \frac{Q ^{2}}{2 C} = \frac{Q ^{2}}{2} [\frac{16}{5 C} - \frac{1}{C}] = \frac{11 Q ^{2}}{10 C}$

Physics - Capacitor Question with Solution | TestHub

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