A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to a potential difference 12.0 V. The battery is now disconnected and a dielectric slab (εᵣ=6) is inserted between the plates. The change in its potential energy after inserting the dielectric sla…

Question

A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to a potential difference 12.0 V. The battery is now disconnected and a dielectric slab (εᵣ=6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ×10⁻¹² J.

Accepted Answer

-750. With the battery connected the capacitor stores Uᵢ=1/2CV²=1/2×12.5×10⁻¹²×(12)²=900×10⁻¹² J. Once the battery is disconnected the charge Q is fixed, so write the energy as U=(Q²)/(2C). Inserting the slab raises the capacitance to εᵣ C, so U_f=(Uᵢ)/(εᵣ)=(900)/6=150×10⁻¹² J. Δ U=U_f-Uᵢ=150-900=-750, i.e. -750×10⁻¹² J. The energy drops because the slab is pulled in by the field, and that work comes out of the stored energy.

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