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: - Dipole consists of charges \( +q \) and \( -q \) separated by distance \( 2a \) - Dipole moment \( \vec{p} = q \cdot 2a \) in the direction from \( -q \) to \( +q \) Let us calculate the electric potential \( V \) at a point at distance \( x \) from the centre of dipole, where \( x \gg a \) \underline{(I) Along the Axis:} The point lies along the axial line (line joining the two charges). Let the positive charge be at \( x = +a \), and the negative charge at \( x = -a \). Then the potential at point \( P \), at distance \( x \) from centre, is: \[ V = \frac{1}{4\pi \varepsilon_0} \left( \frac{+q}{x - a} - \frac{q}{x + a} \right) \] Using binomial expansion for \( x \gg a \): \[ \frac{1}{x \pm a} \approx \frac{1}{x} \left( 1 \mp \frac{a}{x} \right) \Rightarrow V \approx \frac{1}{4\pi \varepsilon_0} \cdot \frac{2qa}{x^2} \] Since \( p = 2aq \), we get: \[ \boxed{V_{\text{axis}} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p}{x^2}} \] \underline{(II) Along the Bisector (Equatorial Line):} The point lies on the perpendicular bisector of the dipole. The distances from both charges are approximately equal: - Distance from each charge: \( r = \sqrt{x^2 + a^2} \approx x \) (since \( x \gg a \)) - The potential at point \( P \) due to both charges is: \[ V = \frac{1}{4\pi \varepsilon_0} \left( \frac{+q}{\sqrt{x^2 + a^2}} - \frac{q}{\sqrt{x^2 + a^2}} \right) = 0 \] Therefore, \[ \boxed{V_{\text{equator}} = 0} \] % Final Answer Statement Answer:
- On the axial line: \( V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p}{x^2} \)
- On the bisector (equatorial line): \( V = 0 \)