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

Solution:

We are given two charges: <math>q_1 = 5 \times 10^{-8} \,\mathrm{C}</math> and <math>q_2 = -3 \times 10^{-8} \text{ C}</math>. The distance between them is <math>d = 16 \text{ cm} = 0.16 \text{ m}</math>.

We need to find the point(s) on the line joining these charges where the net electric potential is zero. The electric potential at any point due to a charge is given by <math>V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r}</math>, where <math>q</math> is the charge and <math>r</math> is the distance from the charge.

Case 1: Point P is between the two charges.

Let P be a point at a distance <math>r</math> from <math>q_1</math>. Then, the distance of P from <math>q_2</math> is <math>(d-r)</math>. The net potential at P is the sum of potentials due to <math>q_1</math> and <math>q_2</math>.

<math display="block">V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1}{r} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_2}{(d-r)}</math>

For the potential to be zero (<math>V = 0</math>):

<math display="block">\frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1}{r} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_2}{(d-r)} = 0</math>

This implies:

<math display="block">\frac{q_1}{r} = -\frac{q_2}{(d-r)}</math>

Substituting the values:

<math display="block">\frac{5 \times 10^{-8}}{r} = -\frac{(-3 \times 10^{-8})}{(0.16-r)}</math>

Simplifying the equation:

<math>5(0.16 - r) = 3r</math>

<math>0.8 - 5r = 3r</math>

<math>0.8 = 8r</math>

<math>r = \frac{0.8}{8} = 0.1 \text{ m} = 10 \text{ cm}</math>

So, the potential is zero at a distance of 10 cm from the positive charge (<math>q_1</math>), between the two charges.

Case 2: Point P is outside the two charges.

Let P be a point at a distance <math>s</math> from the negative charge (<math>q_2</math>). Since <math>q_1</math> is positive and <math>q_2</math> is negative, the point of zero potential must be closer to the charge with the smaller magnitude, which is <math>q_2</math>. Thus, P will be on the side of <math>q_2</math> away from <math>q_1</math>. The distance of P from <math>q_1</math> will be <math>(s+d)</math>.

The net potential at P is:

<math display="block">V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1}{(s+d)} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_2}{s}</math>

For the potential to be zero (<math>V = 0</math>):

<math display="block">\frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1}{(s+d)} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_2}{s} = 0</math>

This implies:

<math display="block">\frac{q_1}{(s+d)} = -\frac{q_2}{s}</math>

Substituting the values:

<math display="block">\frac{5 \times 10^{-8}}{(s+0.16)} = -\frac{(-3 \times 10^{-8})}{s}</math>

Simplifying the equation:

<math>5s = 3(s+0.16)</math>

<math>5s = 3s + 0.48</math>

<math>2s = 0.48</math>

<math>s = \frac{0.48}{2} = 0.24 \text{ m} = 24 \text{ cm}</math>

This distance <math>s</math> is measured from <math>q_2</math>. The distance from <math>q_1</math> would be <math>s+d = 24 + 16 = 40 \text{ cm}</math>.

Therefore, the electric potential is zero at two points: 10 cm from the positive charge between the charges, and 40 cm from the positive charge (or 24 cm from the negative charge) outside the system of charges.

Solution:

We have a regular hexagon with side length <math>l = 10 \text{ cm} = 0.1 \text{ m}</math>. There are six charges, each of magnitude <math>q = 5 \mu C = 5 \times 10^{-6} C</math>, placed at each vertex.

In a regular hexagon, the distance from the center to each vertex is equal to the side length of the hexagon. Therefore, the distance of each vertex from the center (let's call it O) is <math>d = l = 0.1 \text{ m}</math>.

The electric potential at the center O is the sum of the potentials due to each of the six charges. Since all charges are identical and equidistant from the center, we can use the principle of superposition.

The potential due to a single charge <math>q</math> at a distance <math>d</math> is given by <math>V_{single} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{d}</math>.

Since there are six such charges, the total potential at the center is:

<math display="block">V_{total} = 6 \times V_{single} = 6 \times \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{d}</math>

We know that the electrostatic constant <math>\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}</math>.

Substituting the given values:

<math display="block">V_{total} = 6 \times (9 \times 10^9 \text{ Nm}^2\text{C}^{-2}) \times \frac{5 \times 10^{-6} \text{ C}}{0.1 \text{ m}}</math>

Calculating the value:

<math>V_{total} = 6 \times 9 \times 10^9 \times 5 \times 10^{-5} \text{ V}</math>

<math>V_{total} = 270 \times 10^4 \text{ V} = 2.7 \times 10^6 \text{ V}</math>

Therefore, the potential at the centre of the hexagon is <math>2.7 \times 10^6 \text{ V}</math>.