NCERT Solutions For Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

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NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance are provided in this article. The chapter provides good weightage to derivations and numerical problems related to the concepts covered in the chapter. The NCERT Solutions for Class 12 Physics Chapter 2 covers concepts of electrostatic potential, equipotential surfaces, parallel plate capacitors, etc.

The derivation of topics like potential due to an electric dipole, energy stored in the capacitor and potential energy of the system of charges, is frequently asked in the examination. Numerical problems based on the concepts of the effective capacitance of a combination of capacitors are asked regularly in the exams.

Download PDF: NCERT Solutions for Class 12 Physics Chapter 2

NCERT Solutions for Class 12 Physics Chapter 2

NCERT Solutions for Electrostatic Potential and Capacitance are as given below –

Electrostatic Potential and Capacitance Important Topics

Electrostatic Potential is the amount of work done to move a unit charge from a reference point to a specific point inside the electric field without producing an acceleration.

The electrostatic potential of the system is given by the formula:

U = 1/(4πεº) × [q₁q₂/d]

Capacitance is the ratio of change in the electric charge of a system, to the corresponding change in the electric potential.

The formula for capacitance is given by: $\begin{array}{l}C=\frac{Q}{V}\end{array}$

Energy stored in a capacitor: Once the opposite charges are placed on either side of a parallel-plate capacitor, the charges can be allowed to move towards each other through a circuit.

The total energy extracted from a fully charged capacitor is given by the following equation:

$\begin{array}{l}U=\frac{1}{2}CV^2\end{array}$

Electrostatic Potential of a Charge: When a charge, q, is placed in an electric field E, it experiences a force proportional to the magnitude of the charge equal to q × E. If the resultant work done is then divided by the magnitude of charge, it becomes independent of the charge.

The work done by an external force in bringing a unit positive charge from a point A to point B is given by,

$V_B -V_A={U_B-U_A \over q}$

Also Read:

Chapter 2 Related Articles
Van De Graff Generator	Potential due to a point charge	Potential due to an electric dipole
Combination of capacitors	Dielectric Constant	Types of Capacitor
Electrolytic Capacitor	Derivation of Potential Energy	Relation between Electric Field and Electric Potential

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CBSE CLASS XII Related Questions

1.
Read the following paragraph and answer the questions that follow.
A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.
2.
Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.
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3.
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- $ M_x>(M_y + M_z) $
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4.
With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.
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Read the following paragraph and answer the questions that follow.
In an experiment with convex lens of focal length f, the screen is fixed at a distance D from the object. A student slowly moves the lens away from the object towards the screen and finds that she is able to form sharp image of the object for two positions of the lens. The distance between these two positions of the lens is d.
6.
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