Question

An electric dipole of dipole moment \( \vec{p} \) consists of point charges \( +q \) and \( -q \), separated by distance \( 2a \). Derive an expression for the electric potential in terms of its dipole moment at a point at a distance \( x \, (x \gg a) \) from its centre and lying:
(I) along its axis, and
(II) along its bisector (equatorial) line.

Hint:

Electric potential is a scalar quantity. On the axial line, potentials due to the two charges add up, while on the equatorial line, they cancel each other out due to symmetry.

Answer 1

Given: - A dipole comprises charges \( +q \) and \( -q \) separated by a distance \( 2a \). - The dipole moment is defined as \( \vec{p} = q \cdot 2a \), directed from \( -q \) to \( +q \). Calculate the electric potential \( V \) at a point located at a distance \( x \) from the dipole's center, under the condition \( x \gg a \). (I) Along the Axis: The point is situated on the axial line (the line connecting the two charges). Assume the positive charge is at \( x = +a \) and the negative charge is at \( x = -a \). The potential at point \( P \), at a distance \( x \) from the center, is given by: \[V = \frac{1}{4\pi \varepsilon_0} \left( \frac{+q}{x - a} - \frac{q}{x + a} \right)\] Applying the binomial expansion for \( x \gg a \): \[\frac{1}{x \pm a} \approx \frac{1}{x} \left( 1 \mp \frac{a}{x} \right)\] \[\Rightarrow V \approx \frac{1}{4\pi \varepsilon_0} \cdot \frac{2qa}{x^2}\] Substituting \( p = 2aq \): \[\boxed{V_{\text{axis}} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p}{x^2}}\] (II) Along the Bisector (Equatorial Line): The point is located on the perpendicular bisector of the dipole. The distances from both charges to this point are approximately equal: - Distance from each charge: \( r = \sqrt{x^2 + a^2} \approx x \) (since \( x \gg a \)) - The total potential at point \( P \) due to both charges is: \[V = \frac{1}{4\pi \varepsilon_0} \left( \frac{+q}{\sqrt{x^2 + a^2}} - \frac{q}{\sqrt{x^2 + a^2}} \right) = 0\] Consequently, \[\boxed{V_{\text{equator}} = 0}\] Answer: - On the axial line: \( V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p}{x^2} \) - On the bisector (equatorial line): \( V = 0 \)