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Home » Derive an expression for the potential at a point due to a short dipole. Hence show what will be the potential at an axial and an equatorial point :
Derive an expression for the potential at a point due to a short dipole. Hence show what will be the potential at an axial and an equatorial point :
Physics | Previous Year Question Papers
Derive an expression for the potential at a point due to a short dipole. Hence show what will be the potential at an axial and an equatorial point :

Consider an electric dipole A B in which point charges +q and -q are separated by a distance 2l. Let \mathrm{P} be a point on it’s axis seperated by a distance r from the dipole’s center \mathrm{O}.
Electric dipole moment, \mathrm{p}=\mathrm{q}(\mathrm{2l})
Electric potential \left(\mathrm{V}_{A}\right) due to +\mathrm{q} charge

    \[\mathrm{V}_{A}=\frac{1}{4 \pi \varepsilon)} \frac{\mathrm{q}}{\mathrm{r}-1}\]

Electric potential \left(\mathrm{V}_{B}\right) due to +\mathrm{q} charge

    \[\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{-\mathrm{q}}{\mathrm{r}+1}\]

Therefore, total electric potential at P due to dipole will be

    \[\begin{array}{l} \mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\left[\mathrm{~d} \frac{1}{\mathrm{r}-1}-\frac{1}{\mathrm{r}+1}\right] \\ =\frac{\mathrm{q}}{4 \pi \varepsilon_{0}} \frac{21}{\mathrm{r}^{2}-1^{2}} \\ =\frac{\mathrm{p}}{4 \pi \varepsilon_{0}} \frac{1}{\mathrm{r}^{2}-1^{2}} \end{array}\]

Now, 1<\mathrm{r} (short dipole)
\mathrm{V}=\frac{\mathrm{p}}{4 \pi \varepsilon_{0}} \frac{1}{\mathrm{r}^{2}} (for axial position)
For equatorial position, \mathrm{V}=0 as charges are equal in magnitude with opposite signs they cancel each other’s effect.

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