Question

An electric dipole consists of two point charges \(+q\) and \(-q\) separated by a distance \(2a\). Derive an expression for the electric field \(\vec E\) due to this dipole at a point distant \(r\) from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. \(r \gg a\).

Hint:

Electric field due to a dipole: On axial line: \[ E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} \] On equatorial line: \[ E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} \] The equatorial field is opposite to the direction of the dipole moment.

Answer 1

Concept: An electric dipole consists of two equal and opposite charges separated by a small distance. The electric dipole moment is defined as \[ \vec p = q(2a)\,\hat{i} \] and its magnitude is \[ p=2aq. \] The electric field due to a dipole depends upon the position of the observation point. Here we are required to determine the electric field at a point on the equatorial line (perpendicular bisector) of the dipole.

Step 1: Choose a suitable coordinate system. Let the dipole be placed along the \(x\)-axis with \[ +q \text{ at } (a,0) \] and \[ -q \text{ at } (-a,0). \] Let \(P\) be a point on the equatorial line at distance \(r\) from the centre \(O\). The coordinates of \(P\) are \[ (0,r). \]

Step 2: Calculate distance of point \(P\) from each charge. Distance from \(+q\) to \(P\) is \[ d=\sqrt{r^2+a^2}. \] Similarly, distance from \(-q\) to \(P\) is also \[ d=\sqrt{r^2+a^2}. \] Hence both charges are equidistant from the observation point.

Step 3: Determine electric field due to each charge. Magnitude of electric field due to either charge is \[ E_0=\frac{1}{4\pi\varepsilon_0} \frac{q}{r^2+a^2}. \] The field due to \(+q\) is directed away from \(+q\), while the field due to \(-q\) is directed towards \(-q\).

Step 4: Resolve the electric fields into components. Let \(\theta\) be the angle made by the line joining the charge to point \(P\) with the equatorial axis. Then \[ \cos\theta = \frac{a}{\sqrt{r^2+a^2}}. \] The vertical components of the two electric fields are equal in magnitude and opposite in direction. Hence they cancel each other completely. \[ E_y=0. \] The horizontal components are in the same direction and therefore add together.

Step 5: Calculate resultant electric field. Horizontal component due to one charge is \[ E_0\cos\theta. \] Therefore, \[ E = 2E_0\cos\theta. \] Substituting the values, \[ E = 2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2+a^2} \right) \left( \frac{a}{\sqrt{r^2+a^2}} \right). \] Hence \[ E = \frac{1}{4\pi\varepsilon_0} \frac{2aq}{(r^2+a^2)^{3/2}}. \] Since \[ p=2aq, \] we obtain \[ \boxed{ E = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2+a^2)^{3/2}} } \] The direction of this field is opposite to the direction of the dipole moment. Therefore, vectorially \[ \boxed{ \vec E = -\frac{1}{4\pi\varepsilon_0} \frac{\vec p}{(r^2+a^2)^{3/2}} } \]

Step 6: Electric field at a far off point \((r \gg a)\). For a distant point, \[ r^2+a^2 \approx r^2. \] Therefore, \[ (r^2+a^2)^{3/2} \approx r^3. \] Substituting into the above expression, \[ \boxed{ \vec E = -\frac{1}{4\pi\varepsilon_0} \frac{\vec p}{r^3} } \] or in magnitude form, \[ \boxed{ E = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} } \] directed opposite to the dipole moment.