Question

Derive an expression for intensity of electric field at a point broadside position or an equatorial line of an electric dipole.

Answer 1

Hint – In order to solve this problem draw the diagram of the dipoles and use the formula of electric field for a point of charge. Doing this will solve your problem.
Formula used – E = $\dfrac{q}{{4\pi { \in _0}{r^2}}}$

Complete Step-by-Step solution:

Now, suppose that the point P is situated on the right-bisector of the dipole AB at a distance r metre from its mid-point 0 (Fig. (a)]
Again,
Let ${E_1}$​ and ${E_2}$ be the magnitudes of the intensities of the electric field at P due to the charges +q and −q of the dipole respectively.
The distance of P from each charge is $\sqrt {{r^2} + {l^2}} $​.
Therefore,
​${E_1} = \dfrac{{1(q)}}{{(4\pi { \in _0})({r^2} + {l^2})}}$ away from +q
The magnitudes of ${E_1}$ and ${E_2}$​ are equal (but directions are different).
On resolving ${E_1}$ and ${E_2}$ into two components parallel and perpendicular to AB, the components perpendicular to AB (${E_1}\sin \theta $ and ${E_2}\sin \theta $) cancel each other
(because they are equal and opposite), while the components parallel to AB
(${E_1}\cos \theta $ and ${E_2}\cos \theta $), being in the same direction, add up [Fig. (b)]. Hence the resultant intensity of electric field at the point P is
E =${E_1}\cos \theta $ + ${E_2}\cos \theta $
E = $\dfrac{{1(q)\cos \theta }}{{(4\pi { \in _0})({r^2} + {l^2})}} + \dfrac{{1(q)\cos \theta }}{{(4\pi { \in _0})({r^2} + {l^2})}}$
E =$\dfrac{{2q\cos \theta }}{{(4\pi { \in _0})({r^2} + {l^2})}}$
But 2ql=p (moment of electric dipole)
Then ​E = $\dfrac{p}{{(4\pi { \in _0}){{({r^2} + {l^2})}^{\dfrac{3}{2}}}}}$
The direction of electric field E is 'antiparallel' to the dipole axis.
If r is very large compared to $2l\,(r > > 2l)$, then ${l^2}$ may be neglected in comparison to ${r^2}$.
E= $\dfrac{p}{{4\pi { \in _0}{r^3}}}$ Newton/Coulomb is the required electric field.

Note – To solve this problem we need to know that the electric field due to a point charge at a distance r from the charge can be written as $\dfrac{q}{{4\pi { \in _0}{r^2}}}$. Here we have resolved the components of the electric field and then add up the electric field’s components in the direction of the resultant vector and get the value of the electric field due to the dipole. As dipole contains +q and –q so due to that electric field in opposite directions cancels out since they are of equal magnitude but of opposite signs. Knowing all these things will solve your problem.