CBSE Class 12 Physics Chapter 4: Electric Potential and Capacitance NCERT Solutions

Let the two charges be <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 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 = V_1 + V_2 = \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>

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

Substituting the values:

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

<math display="block">\Rightarrow \frac{5}{r} = \frac{3}{(0.16-r)}</math>

<math display="block">\Rightarrow 5(0.16 - r) = 3r</math>

<math display="block">\Rightarrow 0.8 - 5r = 3r</math>

<math display="block">\Rightarrow 0.8 = 8r</math>

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

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

Case 2: Point P is outside the charges.

Since the charges have opposite signs, the potential can also be zero at a point outside the segment joining the charges. Let P be a point at a distance <math>s</math> from the negative charge <math>q_2</math>, such that it is on the line extending from <math>q_1</math> through <math>q_2</math>. The distance of P from <math>q_1</math> will be <math>(d+s)</math>. However, it is more convenient to consider the point P at a distance <math>s</math> from the positive charge <math>q_1</math>, outside the segment, on the side of <math>q_2</math>. In this case, the distance from <math>q_2</math> is <math>(s-d)</math>.

The net potential at P is:

<math display="block">V = V_1 + V_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1}{s} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_2}{(s-d)}</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} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_2}{(s-d)} = 0</math>

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

Substituting the values:

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

<math display="block">\Rightarrow \frac{5}{s} = \frac{3}{(s - 0.16)}</math>

<math display="block">\Rightarrow 5(s - 0.16) = 3s</math>

<math display="block">\Rightarrow 5s - 0.8 = 3s</math>

<math display="block">\Rightarrow 2s = 0.8</math>

<math display="block">\Rightarrow s = \frac{0.8}{2} = 0.4 \text{ m} = 40 \text{ cm}</math>

Thus, the potential is also zero at a point 40 cm from the positive charge <math>q_1</math>, outside the system of charges.

Answer: The electric potential is zero at two points: 10 cm from the positive charge between the two charges, and 40 cm from the positive charge outside the system of charges.

A regular hexagon has 6 vertices. Let the side length of the hexagon be <math>l = 10 \text{ cm} = 0.1 \text{ m}</math>. A charge of <math>q = 5 \mu C = 5 \times 10^{-6} C</math> is placed at each vertex.

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

The electric potential at the center of the hexagon is the sum of the potentials due to the charges at all six vertices. Using the principle of superposition, the total potential <math>V</math> at the center is:

<math display="block">V = V_1 + V_2 + V_3 + V_4 + V_5 + V_6</math>

Since all charges are equal (<math>q</math>) and the distance from each vertex to the center is the same (<math>d</math>), the potential due to each charge at the center is <math>\frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{d}</math>.

Therefore, the total potential is:

<math display="block">V = 6 \times \left( \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{d} \right)</math>

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

Substituting the values:

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

<math display="block">V = 6 \times \left( 9 \times 10^9 \times 50 \times 10^{-6} \right) \text{ V}</math>

<math display="block">V = 6 \times (450 \times 10^3) \text{ V}</math>

<math display="block">V = 2700 \times 10^3 \text{ V} = 2.7 \times 10^6 \text{ V}</math>

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