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.
    Nuclides with the same number of neutrons are called:

      • Isobars
      • Isotones
      • Isotopes
      • Isomers

    • 2.
      In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.


        • 3.
          Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

            • 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.

          • 4.
            A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is \( \sqrt{2} \), and \( \sqrt{3} \).


              • 5.
                The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

                  • P
                  • Q
                  • R
                  • S

                • 6.
                  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
                  CBSE CLASS XII Previous Year Papers

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