Calculate the potential at P due to the charge configuration as shown in the following figure.
If r >> a, then how will you modify the result?

Solution

The charge configuration is as shown in the above fig.
Potential at P is the sum of the potentials due to charges –q, +q and +q which, are at a distance r+a, r and r-a respectively from point P.

∴

V = \frac{1}{4 {πε}_{0}} \frac{- q}{(r + a)} + \frac{1}{4 {πε}_{0}} \frac{q}{r} + \frac{1}{4 {πε}_{0}} \frac{q}{r - a}

\Rightarrow

V = \frac{1}{4 {πε}_{0}} (\frac{q}{r} + \frac{2 qa}{r^{2} - a^{2}})

if,

r > > a,

then

r^{2} - a^{2} \approx r^{2}

∴

V = \frac{1}{4 {πε}_{0}} (\frac{q}{r} + \frac{2 qa}{r^{2}})

Note: When r >> a, the potential at P is simply the potential of a dipole and an isolated charge at distance r.

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Calculate the potential at P due to the charge configuration as shown in the following figure.
If r >> a, then how will you modify the result?