NCERT Solutions for Class 12 Physics Chapter 2: Electrostatic Potential & 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.
A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
the magnetic dipole moment of the coil, and
the magnitude of the counter torque that must be applied to prevent the coil from turning.

View Solution

2.
A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

View Solution

3.
Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side \( a \), kept on a table and carry equal current \( I \). The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude:

\( \dfrac{\mu_0 I}{\pi a} \) and directed along OC
\( \dfrac{\mu_0 I}{\pi a \sqrt{2}} \) and directed along OD
\( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \) and directed along OB
\( \dfrac{2\mu_0 I}{\pi a} \) and directed along OA

View Solution

4.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

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

5.
Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

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

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 \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]

View Solution

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

2.
A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

4.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

5.
Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

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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

2. A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

4.Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

5.Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

4.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

5.
Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.