Question

A capacitor of capacitance c is charged by connecting it to a battery of emf

e. The capacitor is now disconnected and reconnected to the battery with polarity reversed. What is the heat developed in connected wire

Answer 1

Heat developed in connected wire

Answer 2

initial voltage of capacitor , $\xi_0$ = -E [ here negative sign indicates that charge flows through circuit to capacitor because we know, charge flows through positive terminal to negative terminal ]

voltage of capacitor at steady state, $\xi$ = E, [ at steady state charge doesn't flowfrom circuit to capacitor because both are in same potential. e.g., E ]

now, heat developed in connected wire = 1/2 CV²

here C is capacity of capacitance and V is change in voltage .

so, V = E - (-E) = 2E

and hence, heat developed = 1/2 C(2E)² = 2CE²