Introducing a dielectric slab (dielectric constant $K$) fully into a parallel-plate capacitor increases its capacitance by a factor of:

When a dielectric is inserted into an isolated (disconnected) charged capacitor, the voltage across the plates:

The dielectric constant of vacuum is exactly 1.

A 10 μF capacitor is charged to 100 V. Energy stored is $x$ mJ. Find $x$.

Two capacitors $C_1 = 2\,\mu F$ and $C_2 = 3\,\mu F$ are connected in parallel. The equivalent capacitance is:

The capacitance of a parallel-plate capacitor doubles if: