Question

The arrangement of charges shown in FIGURE P25.68 is called a linear electric quadrupole. The positive charges are located at$y=\pm s$. Notice that the net charge is zero. Find an expression for the electric potential on the y-axis at distances$y>>s$. Give your answer in terms of the quadrupole moment,$Q=2q{s}^2$.

Short answer

The expression for electric potential on$y-axis$at distance$y>>s$is$\frac{kQ}{y^3}$

Step-by-step solution

Given Information

We have given that the positive charges are located at$y=\pm s$.

We need to find that an expression for the electric potential on the $y-axis$at distances$y>>s$.

Simplification

As the electric potential is a scalar quantity, we can write the potential at point $y$on the $y-axis$will be given as

$V=k\frac{q}{r}$,

which when considering the distances and charges

$V=kq\frac{1}{y-s}+\frac{1}{y+s}-\frac{2}{y}$

And further expanding this expression, we will get

$V=kq\frac{y(y-s)+y(y-s)-2(y+s)(y-s)}{y(y-s)(y+s)}$

One ought not to panic, as the following can be simplified as

$V=kq\frac{y^2+ys+y^2-ys-2y^2+2ys-2sy+2s^2}{y(y-s)(y+s)}$$=kq\frac{2s^2}{y(y-s)(y+s)}$

Now, let us make the simplification that for $y>>s$,

$y(y-s)(y+s)\approx y^3$,

and using the definition of the quadrupole moment $Q=2q{s}^2$, we can express the result as

$V=\frac{kQ}{y^3}$