Question

Derive an expression for the electric field at a point on the equatorial plane of an electric dipole.

Answer 1

Consider an electric dipole AB consisting charges +q and -q separated by distance 2l and pole strength q. Let, P be the point on the equatorial line at distance d from mid-point of the dipole i.e. O as shown in figure 1.

Now, the electric field, $E_1$ at point P due to north pole along AP is,

$E_1=\frac{1}{4\pi {ϵ}_0}\frac{q}{A{P}^2}$

$E_1=\frac{1}{4\pi {ϵ}_0}\frac{q}{(d^2+l^2{)}^2}$ .........(1) ($\because A{P}^2=O{A}^2+O{P}^2$)

Similarly, the electric field, $E_2$ at point P due to south pole along PB is,

$E_2=\frac{1}{4\pi {ϵ}_0}\frac{q}{P{B}^2}$

$E_2=\frac{1}{4\pi {ϵ}_0}\frac{q}{(d^2+l^2{)}^2}$ .........(2) ($\because P{B}^2=O{B}^2+O{P}^2$)

From equations (1) and (2), we get

$E_1=E_2=E\left(say\right)$

Now, the net electric field $E_P$ at P due to the dipole is

$E_P=E_1+E_2=2E$ ........................(3)

Resolving $E_1$ and $E_2$ into horizontal and vertical components, as shown in figure 2. The vertical components will be canceled and horizontal will be added together. So, we can write

$E_P=2E\cos \theta$ .............from(3)

$E_P=2\left(\frac{1}{4\pi {ϵ}_0}\frac{q}{(d^2+l^2{)}^2}\right)\left(\frac{l}{\sqrt{(}{d}^2+l^2)}\right)$ ............($\because \cos \theta =\frac{l}{\sqrt{(}{d}^2+l^2)}$)

For a short dipole, $l<<d$, therefore

$B_P=\frac{1}{4\pi {ϵ}_0}\frac{2ql}{d^3}$

$B_P=\frac{1}{4\pi {ϵ}_0}\frac{Q}{d^3}$ ........................($\because Q=2ql$)