91Ó°ÊÓ

An electric dipole at the origin consists of two charges q spaced apart along the y-axis.

(a).

Take an arbitrary point in the xy plane with the position vector with respect to the origin.

$\stackrel{→}{r}=x\stackrel{→}{i}+y\stackrel{→}{j}$

Set s=2 a. The position of this point with respect to the +q charge is

${\stackrel{→}{r}}_1=(x-a)\stackrel{→}{i}+y\stackrel{→}{j}$

And the position of this point with respect to the -q charge is

${\stackrel{→}{r}}_2=(x+a)\stackrel{→}{i}+y\stackrel{→}{j}$

The potential due to +q is then equal to

$V_1=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q}{r_1}$

and the potential due to -q is

$V_2=-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q}{r_2}$

We havethat

$r_1=\sqrt{(x-a{)}^2+y^2},r_2=\sqrt{(x+a{)}^2+y^2}$

The total potential due to both charges is

$V(x,y)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{\sqrt{(x-a{)}^2+y^2}}-\frac{1}{\sqrt{(x+a{)}^2+y^2}}\right)$

Remember that a=s/2 so

$V(x,y)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{\sqrt{{\left(x-\frac{s}{2}\right)}^2+y^2}}-\frac{1}{\sqrt{{\left(x+\frac{s}{2}\right)}^2+y^2}}\right)$

(b).

The binomial approximation says

$(1+x)^{\alpha} \approx 1+\alpha x$, for $x \ll 1$

Rewrite

$$

\begin{aligned}

\frac{1}{\sqrt{(x-a)^{2}+y^{2}}} &=\left((x-a)^{2}+y^{2}\right)^{-\frac{1}{2}}=\left(x^{2}+y^{2}-2 a x+a^{2}\right)^{-\frac{1}{2}} \\

&=\left[\text { neglect } a^{2} \text { compared to other terms }\right]=\\

&=\left(x^{2}+y^{2}\right)^{-\frac{1}{2}}\left(1-\frac{2 x}{x^{2}+y^{2}} a\right)^{-\frac{1}{2}}=\\

&=[\text { apply binomial approximation }]=\\

&=\left(x^{2}+y^{2}\right)^{-\frac{1}{2}}\left(1+\frac{x a}{x^{2}+y^{2}}\right)

\end{aligned}

$$

Similarly

$$

\begin{aligned}

\frac{1}{\sqrt{(x+a)^{2}+y^{2}}} &=\left((x+a)^{2}+y^{2}\right)^{-\frac{1}{2}}=\left(x^{2}+y^{2}+2 a x+a^{2}\right)^{-\frac{1}{2}} \\

&=\left[\text { neglect } a^{2} \text { compared to other terms }\right]=\\

&=\left(x^{2}+y^{2}\right)^{-\frac{1}{2}}\left(1+\frac{2 x}{x^{2}+y^{2}} a\right)^{-\frac{1}{2}}=\\

&=[\text { apply binomial approximation }]=\\

&=\left(x^{2}+y^{2}\right)^{-\frac{1}{2}}\left(1-\frac{x a}{x^{2}+y^{2}}\right)

\end{aligned}

$$

Therefore

$$

\frac{1}{\sqrt{(x-a)^{2}+y^{2}}}-\frac{1}{\sqrt{(x+a)^{2}+y^{2}}}=\frac{2 x a}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}} .

$$

This yields for the potential

$$

V(x, y)=\frac{q}{4 \pi \varepsilon_{0}} \frac{2 x a}{\left(x^{2}+y^{2}\right)^{\frac{2}{2}}}

$$

Remember that $s=2 a$ so

$$

V(x, y)=\frac{q}{4 \pi \varepsilon_{0}} \frac{x s}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}}

$$

We will use the expression for V derived in part b. to find the components of the field. As we know, the x and y components of the field are given by minus the partial derivative of V with respect to x and y.