Physics Class 12 NCERT Solutions: Chapter 2 Electrostatic Potential and Capacitance Part 13 (For CBSE, ICSE, IAS, NET, NRA 2022)

Get unlimited access to the best preparation resource for CBSE/Class-12 : get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-12.

Formula of Electric Field in Capacitor

Q: 28. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to , where is the charge on the capacitor, and is the magnitude of electric field between the plates. Explain the origin of the factor .


Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of . Hence, work done by the force to do so

As a result, the potential energy of the capacitor increases by an amount given as .


u = Energy density

A = Area of each plate

d = Distance between the plates

V = Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.. ,

Electric intensity is given by,

However, capacitance,

Charge on the capacitor is given by,

The physical origin of the factor, in the force formula lies in the fact that just outside the conductor, field is and inside it is zero. Hence, it is the average value, , of the field that contributes to the force.

Q: 29. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports .

Q_29_Spherical Capacitor

Show that the capacitance of a spherical capacitor is given by

Where and are the radii of outer and inner spheres, respectively.


Radius of the outer shell

Radius of the inner shell

The inner surface of the outer shell has charge .

The outer surface of the inner shell has induced charge .

Potential difference between the two shells is given by,


Permittivity of free space

Capacitance of the given system is given by,

Hence, proved.

Developed by: