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#### Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^{−12} F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

#### Solution

#### Similar questions

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery

The plates of a parallel plate capacitor have an area of 90 cm^{2} each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

**(a)** How much electrostatic energy is stored by the capacitor?

**(b)** View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume *u*. Hence arrive at a relation between *u *and the magnitude of electric field *E *between the plates.

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^{−3} m^{2} and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10^{7} Vm^{−1}. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.