(II) Calculate the electric potential due to a dipole whose dipole moment is \({\bf{4}}{\bf{.2 \times 1}}{{\bf{0}}^{{\bf{ - 30}}}}\;{\bf{C \times m}}\) at a point \({\bf{2}}{\bf{.4 \times 1}}{{\bf{0}}^{{\bf{ - 9}}}}\;{\bf{m}}\) away if this point is (a) along the axis of the dipole nearer the positive charge; (b) 45° above the axis but nearer the positive charge; (c) 45° above the axis but nearer the negative charge.

In an electric dipole, two point charges each of magnitude Qare separated by a distance l.

The potential due to a dipole at an arbitrary point is given as:

\(V = \frac{{kp\cos \theta }}{{{r^2}}}\) … (i)

Here, k is the Coulomb’s constant, p is the dipole moment, r is the distance from the dipole.

The dipole moment is,\(p = 4.2 \times {10^{ - 30}}\;{\rm{C}} \cdot {\rm{m}}\).

The distance of the point from the dipole is,\(r = 2.4 \times {10^{ - 9}}\;{\rm{m}}\).

The angle is,\(\theta = 45^\circ \).

Along the axis of dipole near the positive charge,\({\theta _1} = 0^\circ \)

The electric potential due to the dipole is calculated as:

\({V_1} = \frac{{kp\cos {\theta _1}}}{{{r^2}}}\)

Substitute the values in the above expression.

\(\begin{aligned}{V_1} &= \left( {\frac{{\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {4.2 \times {{10}^{ - 30}}\;{\rm{C}} \cdot {\rm{m}}} \right)\cos 0^\circ }}{{{{\left( {2.4 \times {{10}^{ - 9}}\;{\rm{m}}} \right)}^2}}}} \right)\\{V_1} &= 6.56 \times {10^{ - 3}}\;{\rm{V}}\end{aligned}\)

Thus, the electric potential along the axis of the dipole is \(6.56 \times {10^{ - 3}}\;{\rm{V}}\).

The angle above the axis but near the positive charge is, \({\theta _2} = 45^\circ \)

The electric potential due to the dipole is calculated as:

\({V_2} = \frac{{kp\cos {\theta _2}}}{{{r^2}}}\)

Substitute the values in the above expression.

\(\begin{aligned}{V_2} &= \left( {\frac{{\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {4.2 \times {{10}^{ - 30}}\;{\rm{C}} \cdot {\rm{m}}} \right)\cos 45^\circ }}{{{{\left( {2.4 \times {{10}^{ - 9}}\;{\rm{m}}} \right)}^2}}}} \right)\\{V_2} &= 4.6 \times {10^{ - 3}}\;{\rm{V}}\end{aligned}\)

Thus, the electric potential 45° above the axis but near the positive charge is \(4.6 \times {10^{ - 3}}\;{\rm{V}}\).

The angle above the axis but near the negative charge is, \(45^\circ \), so the angle made with the positive charge is, \({\theta _3} = 135^\circ \).

The electric potential due to the dipole is calculated as:

\({V_3} = \frac{{kp\cos {\theta _3}}}{{{r^2}}}\)

Substitute the values in the above expression.

\(\begin{aligned}{V_3} &= \left[ {\frac{{\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {4.2 \times {{10}^{ - 30}}\;{\rm{C}} \cdot {\rm{m}}} \right)\cos 135^\circ }}{{{{\left( {2.4 \times {{10}^{ - 9}}\;{\rm{m}}} \right)}^2}}}} \right]\\{V_3} &= - 4.6 \times {10^{ - 3}}\;{\rm{V}}\end{aligned}\)

Thus, the electric potential 45° above the axis but near the negative charge is \( - 4.6 \times {10^{ - 3}}\;{\rm{V}}\).