Physics Test - ...

A parallel plate capacitor has a capacity 80×10⁻⁶ F when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of 30 V by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is :

Consider the circuit given here. The potential difference VBC between the points B and C is :

Four metallic plates are arranged as shown in the figure. If d is the distance between each plate then capacitance of the given system between points A and B is :

(given  d<<A}

Charge Q taken from the battery of 12V in the circuit is :

The potential difference between the plate of a capacitor is increased by 20%. The energy stored on the capacitor increases by:

Area of each of the conducting plates 1, 2, 3, 4, 5 and 6 is A. The system is kept in air. Find the the capacitance between A and B (1 & 4, 2 & 5 and 3 & 6 are pairs of parallel plates).

The equivalent capacitance between x and y is:

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is-

A parallel plate air capacitor has capacity ′C′ farad, potential ′V′ volt and energy ′E′ joule. When the gap between the plates is completely filled with dielectric.

A parallel plate capacitor with air as a dielectric has capacitance C. A slab of dielectric constant K, having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be: