# derivation of electric potential at a point on axial line of an electric dipole

Suppose P is a point on the axial position of the dipole.
Length of dipole=2a
Suppose point P is at the distance 'r' from the center of the dipole.
We know that potential at a point is given as,
$\mathrm{V}=\frac{1}{4\mathrm{\pi }{\in }_{0}}.\frac{\mathrm{Q}}{\mathrm{r}}$
So, potential at P due to q is,
${\mathrm{V}}_{\mathrm{q}}=\frac{1}{4\mathrm{\pi }{\in }_{0}}.\frac{\mathrm{q}}{\left(\mathrm{a}+\mathrm{r}\right)}$
potential at P due to –q is,
${\mathrm{v}}_{-\mathrm{q}}=\frac{1}{4\mathrm{\pi }{\in }_{0}}.\frac{-\mathrm{q}}{\left(\mathrm{r}-\mathrm{a}\right)}$
Total potential at P is
$\mathrm{V}={\mathrm{V}}_{\mathrm{q}}+{\mathrm{V}}_{-\mathrm{q}}\phantom{\rule{0ex}{0ex}}=\frac{1}{4\mathrm{\pi }{\in }_{0}}.\frac{\mathrm{q}}{\left(\mathrm{a}+\mathrm{r}\right)}+\frac{1}{4\mathrm{\pi }{\in }_{0}}.\frac{-\mathrm{q}}{\left(\mathrm{r}-\mathrm{a}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_{0}}\left[\frac{1}{\left(\mathrm{a}+\mathrm{r}\right)}+\frac{1}{\left(\mathrm{a}-\mathrm{r}\right)}\right]\phantom{\rule{0ex}{0ex}}\mathrm{V}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_{0}}.\frac{2\mathrm{a}}{\left({\mathrm{a}}^{2}-{\mathrm{r}}^{2}\right)}$

• 51

u bdont need 2 kno it jst undrstnd d concept not derivation

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