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

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Jasmine Grover

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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 – 

NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics NCERT Solutions Physics

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:

\(\begin{array}{l}C=\frac{Q}{V}\end{array}\)

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,

\(V_B -V_A={U_B-U_A \over q}\)


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CBSE CLASS XII Related Questions

  • 1.
    A 500 nm photon is incident normally on a perfectly reflecting surface and is reflected. The value of momentum transferred to the surface is:

      • \( 3.87 \times 10^{-43} \, \text{kg} \, \text{ms}^{-1} \)
      • \( 2.5 \times 10^{-30} \, \text{kg} \, \text{ms}^{-1} \)
      • \( 2.65 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)
      • \( 1.33 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)

    • 2.
      Differentiate between inductive reactance, capacitive reactance and impedance of an ac circuit.
      An ideal inductor and an ideal capacitor are connected in series across an ac voltage. Plot a graph showing variation of net reactance of the circuit with frequency of the applied ac voltage.


        • 3.
          Determine the current in the \( 3 \, \Omega \) branch of a Wheatstone Bridge in the circuit shown in the figure.


            • 4.
              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.


                • 5.
                  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.


                    • 6.
                      Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

                        CBSE CLASS XII Previous Year Papers

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