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.
    In a telescope the objective has much larger aperture than the eye piece. Why ?


      • 2.
        A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.


          • 3.
            Read the following paragraph and answer the questions that follow.
            A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.


              • 4.
                This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?


                  • 5.
                    Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.


                      • 6.
                        If T is the time period of the rotation of the coil, at what values of t in a cycle, the emf generator is maximum ?

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