An air capacitor has \(C_0=25\,\mathrm{pF}\). It is completely filled with dielectric of \(\kappa=4.0\). Find the new capacitance.

A \(12\,\mathrm{V}\) battery remains connected to a capacitor while a dielectric with \(\kappa=3.0\) is inserted. By what factor do the capacitance and free charge on the plates change?

An isolated capacitor is completely filled with a dielectric of \(\kappa=5.0\). Determine the factors by which its capacitance, voltage, electric field, and stored energy change.

The same dielectric is inserted into two identical capacitors. Capacitor A remains connected to a battery at fixed \(V\); capacitor B is isolated at fixed \(Q\). Derive the energy change factor for each capacitor and explain why the two results are different.

A \(60\,\mathrm{pF}\) air capacitor is completely filled with a dielectric of \(\kappa=2.5\). Find the new capacitance.

A dielectric-filled parallel-plate capacitor has \(\kappa=3.2\), \(A=0.015\,\mathrm{m^2}\), and \(d=0.75\,\mathrm{mm}\). Find its capacitance.

An isolated capacitor is filled with \(\kappa=4.0\). If its original voltage was \(80\,\mathrm{V}\), find the final voltage.

A capacitor remains connected to a \(24\,\mathrm{V}\) battery while a dielectric of \(\kappa=5.0\) is inserted. If the original charge was \(12\,\mu\mathrm{C}\), find the final charge.

A dielectric-filled parallel-plate capacitor has free surface charge density \(1.8\times10^{-6}\,\mathrm{C\,m^{-2}}\) and \(\kappa=6.0\). Find the electric field inside the dielectric.

An air capacitor has \(C_0=120\,\mathrm{pF}\). A material is inserted and the measured capacitance becomes \(420\,\mathrm{pF}\). Find \(\kappa\). If the capacitor is isolated, by what factor does its voltage change?

A dielectric has breakdown field \(1.5\times10^7\,\mathrm{V\,m^{-1}}\). It fills a parallel-plate capacitor with plate spacing \(0.20\,\mathrm{mm}\). Estimate the maximum safe voltage.

A parallel-plate capacitor is half-filled side-by-side: half the plate area contains dielectric \(\kappa\), and half remains vacuum, with both regions spanning the full separation \(d\). Derive the equivalent capacitance.

A dielectric slab of thickness \(t\) and dielectric constant \(\kappa\) is inserted between parallel plates of area \(A\) and separation \(d\), leaving vacuum in the remaining gap. Derive \(C\), neglecting fringing.

For an isolated capacitor, a dielectric of \(\kappa=3.0\) is inserted. If the original stored energy was \(9.0\,\mathrm{mJ}\), find the final energy and explain the energy change.

For a battery-connected capacitor, a dielectric of \(\kappa=3.0\) is inserted. If the original capacitor energy was \(9.0\,\mathrm{mJ}\), find the final capacitor energy and state whether the battery must supply charge.

A dielectric partially inserted into a battery-connected parallel-plate capacitor is pulled inward. Explain the direction of the force using capacitance and energy, without deriving a full force formula.

A dielectric slab is removed slowly from an isolated capacitor. Determine how \(C\), \(V\), \(E\), and \(U\) change, assuming the dielectric initially filled the gap and had dielectric constant \(\kappa\).

Two identical isolated capacitors carry the same charge. One is filled with dielectric \(\kappa_1\), the other with \(\kappa_2>\kappa_1\). Compare their voltages, fields, and stored energies.

A parallel-plate capacitor has plate separation \(d\). The first half of the gap is filled with dielectric \(\kappa_1\), and the second half with dielectric \(\kappa_2\). Derive the equivalent capacitance and check the case \(\kappa_1=\kappa_2\).

A capacitor of area \(A\) and separation \(d\) is filled with a dielectric whose constant varies linearly across the gap: \(\kappa(x)=\kappa_0(1+\alpha x/d)\), where \(0\le x\le d\). Derive the capacitance, assuming the displacement field is uniform and normal to the plates.