Question

A capacitor with capacitance \(3\times10^{-9}\,\mathrm{F}\) is connected to \(400\,\mathrm{V}\). A dielectric slab with \(K=3\) completely fills the space, and the voltage is kept constant. What is the energy change when the slab is removed?

Options

• The energy decreases by \(48\times10^{-5}\,\mathrm{J}\)

• The energy increases by \(48\times10^{-5}\,\mathrm{J}\)

• The energy does not change

• The energy decreases by \(24\times10^{-5}\,\mathrm{J}\)

Answer 1

At constant voltage, \(U=\frac{1}{2}CV^2\), so the energy is proportional to capacitance. The dielectric-present energy is \(72\times10^{-5}\,\mathrm{J}\), compared with \(24\times10^{-5}\,\mathrm{J}\) with air; removing it decreases the energy by \(48\times10^{-5}\,\mathrm{J}\).