Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2)QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to 1/2QE, where Q is the charge on the capacitor and E is the magnitude of electric field between the plates.

The area of each plate of a parallel plate capacitor is A = 600 cm^2 and their separation is d = 2.0 mm. The capacitor is connected to a 200 V dc source. Find out the uniform electric field between the plates in SI unit?

The electrostatic force between the metal plates of an isotated parallel plate capacitor C having a charge Q and area A, is

A slab of dielectric constant k has the same area as the plates of a parallel plate capacitor but has a thickness 3d/4, where is the separation between the plates. How is the capacitance changed, when the slab is inserted between the plates

Distance between the two plates of a parallel plate capacitor, each having area A, in vecuum is D. Calculate its capacitance.

A slab of material of dielectric constant k has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, when d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

A parallel plate capacitor with oil(K=2) has a capacitance C. If the oil is removed, then the capacity of the capacitor becomes