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.
The resistance of a metal wire at $20^\circ\text{C}$ is $1.05\,\Omega$ and at $100^\circ\text{C}$ is $1.38\,\Omega$. Determine the temperature coefficient of resistivity of this metal.
View Solution
2.
Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.
View Solution
3.
A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:
(I) the formation of maxima and minima in the diffraction pattern, and
(II) why the maxima go on becoming weaker and weaker with its increasing number $n$.
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4.
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 $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:
- 5 V
- 3.5 V
- 2.5 V
- Zero V
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5.
A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:
- 80 Wb
- 16 Wb
- $8 \times 10^{-2}$ Wb
- $8 \times 10^{-3}$ Wb
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6.
Consider a cylindrical conductor of length $l$ and area of cross-section $A$. Current $I$ is maintained in the conductor and electrons drift with velocity $\vec{v}_d$ ($|\vec{v}_d| = \frac{e|\vec{E}|}{m}\tau$), (where symbols have their usual meanings). Show that the conductivity $\sigma$ of the material of the conductor is given by
$$\sigma = \frac{ne^2}{m}\tau.$$
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CBSE CLASS XII Previous Year Papers

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

Applications of Gauss Law: Formula & Gauss Theorem

Coulomb’s Law: Formula, Vector Form & Limitations

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.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

2.
Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

4.
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 $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

5.
A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

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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.The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

2.Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

4.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 $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

5.A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

2.
Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

4.
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 $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

5.
A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is: