Electrostatic Potential

Previous Year Examination Questions

1 Mark Questions

1. The figure shows the field lines of a positive charge. Is the work done by the field is moving

a small positive charge from Q to P positive or negative? [Foreign 2014]

Ans.

Work done by charge is given by

<math>W = q</math> (potential at Q – potential at P).

where, <math>q = \text{small positive charge}</math>

(1/2)

The electric potential at a point distant r due to the field created by a positive charge Q is given by

<math display="block">V = \frac{1 \, q}{4\pi \varepsilon_0 \, r}</math>

<math>\ddot{\cdot}</math> <math>r_p < r_Q</math> <math>\Rightarrow</math> <math>V_p > V_Q</math>

(1/2)

So, work done will be negative.

1. For any charge configuration, equipotential surface through a point is a normal to the electric field. Justify. [Delhi 2014]

Ans.

No work is done in moving the test charge from one point of an equipotential surface to the other. (1/2)

<math display="block">W_B - W_A = 0 = -\int \mathbf{E} \cdot dI</math> <math>\Rightarrow</math> <math>\mathbf{E} \cdot dl = 0</math> Hence, <math>\mathbf{E} \perp dl</math>

(1/2)

1. Two charges <math>2 \mu C</math> and <math>-2 \mu C</math> are placed at points A and B, 5 cm apart. Depict an equipotential surface of the system. [Delhi 2013C]

Ans.

Given, <math>q_A = 2 \,\mu\text{C} = 2 \times 10^{-6} \,\text{C}</math>

<math>q_B = -2 \,\mu\text{C} = -2 \times 10^{-6} \,\text{C}</math>

<math>A \stackrel{\longleftarrow}{\longleftrightarrow} X \stackrel{\longleftarrow}{\longleftrightarrow} (5-X) \stackrel{\longrightarrow}{\longleftrightarrow} B</math> ----5 cm and <math>r = 5</math> cm

<math display="block">\therefore \text{ Potential, } V = \frac{2 \times 10^{-6}}{4\pi \varepsilon_0 \, x \times 10^{-2}}</math> <math display="block">+\frac{-2\times10^{-6}}{4\pi\epsilon_0(5-x)\times10^{-2}}</math>

<math display="block">\therefore \frac{2 \times 10^{-6}}{4\pi \varepsilon_0 x \times 10^{-2}} = \frac{2 \times 10^{-6}}{4\pi \varepsilon_0 (5 - x) \times 10^{-2}} \ [\because V = 0]</math>

<math>x = 5 - x</math> <math>x = 2.5</math> (1)

1. Why electrostatic potential is constant throughout the volume of the conductor and has the same value as on its surface? [Delhi 2012]

Ans. Since, electric field intensity inside the conductor is zero. So, electrostatic potential is a constant.

But, <math display="block">E = -\frac{\Delta V}{\Delta r}</math>

<math display="block">E = 0, \ \Delta V = 0</math>

or <math display="block">V_2 - V_1 = 0, \ V_2 = V_1</math>

1. Why is the potential inside a hollow spherical charged conductor is constant and has the same value as on its surface?[Foreign 2012]

Ans. Electric field inside the hollow spherical charged conductor is zero. So, no work is done in moving a charge inside the shell.

This implies that potential is a constant and therefore, equal to its value at the surface, i.e.

<math display="block">V = \frac{1}{4\pi\,\varepsilon_0} \cdot \frac{q}{R}</math>

1. Why there is no work done in moving a charge from one point to another on an equipotential surface? [Foreign 2012]

Ans.