Question

An electric dipole consists of two charges of equal magnitude and opposite sign separated by distance 2a, as shown in the figure. the dipole is along the x-axis and is centred at the origin.

calculate the electric potential at point P?

calculate v at a point far from the dipole?

Answer 1

Let, dipole of the $K$. $B$ at point where $+a$ charge and $P$ at a point,

$E_1=\frac{9}{K{\left(BP\right)}^2}=\frac{9}{{K(x-a)}^2}$

($E_1$, towards $\overline{OP}$)

and, $A$ at a point $-9$ charge where the electric field at point $P$

$E_2=\frac{9}{K{\left(AP\right)}^2}$

$=\frac{9}{K{(x+a)}^2}$

($E_2$, towards $\overline{PO}$)

Since, $E_1$ and $E_2$ are same line but opposite in direction and $E_1>E_2$ $\therefore$ $P$ at a point,

The resultant of electric field $P$.

$E=E_1-E_2$

$=\frac{9}{K}[\frac{1}{{(x-a)}^2}-\frac{1}{{(x+a)}^2}]$

$=\frac{9}{K}\left[\frac{{(x+a)}^2-{(x-a)}^2}{{(x-a)}^2{(x-a)}^2}\right]$

$E=\frac{9.4xa}{K{(x^2-a^2)}^2}$

$=\frac{2px}{K{(x^2-a^2)}^2}$ $[\because P=2aq]$

Calculation at $V$ at a point four from the dipole

$V=\frac{E}{q}$

$=\frac{2px}{Kq{(x^2-a^2)}^2}$