- #1

- 236

- 16

## Homework Statement

An electric dipole consists of two charges of equal magnitude ##q## and opposite sign, which are kept at a distance ##d## apart. The dipole moment is ##p= qd## .

Let us next place two such dipoles, placed at distance ##r## apart, as shown in the accompanying figure.

a) Assuming that the potential energy for the charges while at infinity is zero, find the exact potential energy of the configuration in terms of ##d, r, q ## and fundamental constants.

b)When ##d<<r## , approximate your previous result in terms of ##p, r ## and fundamental constants.

## Homework Equations

##V = \frac {1}{4\pi \epsilon_0} \frac{q}{r}##

##W = Vq##

## The Attempt at a Solution

For the positve charge of A dipole, I calculated the potential energy

##= +q \cdot \frac{1}{4\pi \epsilon _0} (\frac{-q}{r} + \frac {+q}{\sqrt{r^2 +d^2}} + \frac{-q}{d}) ##, because, ##W = Vq##

The potental energy for the other three charges are the same.

So, the potential energy of the configuration is

## = 4q \cdot \frac{1}{4\pi \epsilon _0} (\frac{-q}{r} + \frac {+q}{\sqrt{r^2 +d^2}} + \frac{-q}{d})##

##= \frac{q^2}{\pi \epsilon _0} (- \frac{1}{r} + \frac {1}{\sqrt{r^2 +d^2}} - \frac{1}{d})##;

Then , I can't find any way how to approximate the result when ##d << r##;

In this case, I substituted ##r^2## for ##r^2 + d^2## ;

So, two of the terms in bracket are cancelled.

Then I plugged in ## q = \frac{p}{d}## ;

But, still ##d## is there.