Electrostatics

The energy stored in a 5 μF capacitor charged to 100 V is:

The space between the plates of a capacitor of capacitance C is completely filled with a dielectric of dielectric constant K. The new capacitance is:

A capacitor is charged by a battery and then disconnected. A dielectric slab of dielectric constant K is then inserted to fill the gap completely. The stored energy U becomes:

A capacitor remains connected to a battery while a dielectric slab of dielectric constant K is inserted to fill the gap completely. The stored energy U becomes:

A dielectric slab of thickness t and dielectric constant K is placed between the plates of a parallel-plate capacitor of plate area A and separation d (t smaller than d). The capacitance becomes:

An air capacitor has a capacitance of 10 pF. Half the gap between its plates (a layer of thickness d/2, covering the full plate area) is filled with a dielectric of K = 5. The new capacitance is about:

An air capacitor has a capacitance of 10 pF. A dielectric of K = 5 fills the full gap over half of the plate area, the other half remaining air. The new capacitance is:

The magnitude of the attractive force between the plates of an isolated parallel-plate capacitor with charge Q and plate area A is:

When a dielectric is placed in an external electric field, the net field inside the dielectric:

A 2 μF capacitor is charged to 200 V and disconnected from the battery. A dielectric slab of K = 4 is then inserted to fill the gap completely. The new potential difference is:

A 4 μF capacitor carries a charge of 40 μC. The energy stored in it is:

An uncharged capacitor is charged by connecting it directly to a battery of emf V through a resistor. The fraction of the energy supplied by the battery that is stored in the capacitor is:

The dielectric strength of a material is:

Find the energy (in millijoule) stored in a 4 μF capacitor charged to 50 V.

The capacitance of an air capacitor is 6 pF. When the space between its plates is completely filled with a dielectric, the capacitance becomes 24 pF. Find the dielectric constant.

A 10 μF capacitor is charged to 100 V and then disconnected from the battery. A dielectric slab of K = 2 is inserted to fill the gap completely. Find the decrease in stored energy, in millijoule.