Question

What is the capacitance of an isolated parallel plate capacitor when a full dielectric slab of dielectric constant \[K\] is inserted?

Options

• \[C=KC_0=\frac{K\varepsilon_0A}{d}\]

• \[C=\frac{C_0}{K}=\frac{\varepsilon_0A}{Kd}\]

• \[C=C_0-K\]

• \[C=K+C_0\]

Answer 1

With charge constant, the dielectric reduces the potential difference by a factor of \[K\]. Since capacitance is \[Q/V\], the capacitance increases to \[C=KC_0\].