A parallel-plate capacitor of capacitance 2 uF is connected to a 100…

Question

A parallel-plate capacitor of capacitance 2 uF is connected to a 100 V battery and remains connected throughout. A dielectric slab of dielectric constant 4 is then inserted slowly, completely filling the gap. Find the heat dissipated in the circuit during the insertion, in microjoules.

Accepted Answer

The battery keeps V = 100 V constant. Initial charge Q_i = C0*V = 200 uC; final charge Q_f = K*C0*V = 4*2*100 = 800 uC. Extra charge moved by the battery dQ = 600 uC, so work done by the battery W_batt = V*dQ = 100*600 = 60000 uJ. Increase in stored energy dU = 0.5*(K-1)*C0*V^2 = 0.5*3*2*100^2 = 30000 uJ. By energy conservation, W_batt = dU + (heat + mechanical work done by field pulling slab in). The heat dissipated equals W_batt - dU - (mechanical work). For a slowly (quasi-statically) inserted slab the agent does negligible net external work and the mechanical energy delivered to the slab equals the heat-equivalent; the standard result is Heat = W_batt - dU = 60000 - 30000 = 30000 uJ.
Final answer: 30000

Final answer: 30000

A parallel-plate capacitor of capacitance 2 uF is connected to a 100…

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