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:

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Van De Graff Generator	Potential due to a point charge	Potential due to an electric dipole
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CBSE CLASS XII Related Questions

1.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Both Assertion (A) and Reason (R) are false.
View Solution
2.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:
- 5 V
- 3.5 V
- 2.5 V
- Zero V
View Solution
3.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.
View Solution
4.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –
- \((P_1+P_2)\)
- \(\dfrac{P_1+P_2}{2}\)
- \(\dfrac{P_1P_2}{2(P_1+P_2)}\)
- \(\dfrac{P_1P_2}{4(P_1+P_2)}\)
View Solution
5.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.
View Solution
6.
Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.
View Solution

View All

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Electric Charges and Fields: Flux, Gauss Law & Coulomb's Law

Applications of Gauss Law: Formula & Gauss Theorem

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Electric Charge: Definition, Formula, Types and Properties

Electric Field: Definition, Formula & Electric Field Direction

Gauss Law: Formula, Gauss Theorem & Electric Flux

Dipole in a Uniform External Field: Torque, Frequency, and Time Period

Electric Dipole: Definition, Significance and Electric Dipole Moment

Electric Flux: Formula, Equation, Symbol & SI Unit

Electrostatic Potential: Definition, Formula and SI Unit

Electric Potential Due to a Point Charge: Derivation & Formula

Potential Due to an Electric Dipole: Introduction, Formula and Derivation

Equipotential Surface: Work Done, Properties & Magnitude

Van de Graaff Generator: Principle, Working, and Construction

Energy Stored in a Capacitor: Formula, Derivation and Applications

Electrostatic Potential and Capacitance: Introduction & Derivations

Electric Charges and Fields Important Questions

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

NCERT Solutions for Class 12 Physics Chapter 2

Electrostatic Potential and Capacitance Important Topics

CBSE CLASS XII Related Questions

1.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

2.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

3.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

5.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

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

NCERT Solutions for Class 12 Physics Chapter 2

Electrostatic Potential and Capacitance Important Topics

CBSE CLASS XII Related Questions

1.Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

2.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

3.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

5. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

2.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

3.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

5.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.