Limit at infinity of an electric quadrupole

[Monocles]

This is for my intro physics 2 class

Homework Statement

Consider the charges Q at (-a, 0), -2Q at (0, 0) and Q at (a, 0). Such a combination of charges, with zero net charge and with zero net dipole moment, is called an electric quadrupole. a. Find the electric field along the x acis, for x > a. b. Show that, for x >> a, the electric field varies by x^-4. Find the coefficient.

Homework Equations

\vec{E} = \frac{kQ}_{r^{2}}\hat{r}

Here is the answer I got for part a which was correct.

\vec{E} = kQ\left[\frac{1}_{(x+a)^{2}}} + \frac{1}_{(x-a)^{2}}} - \frac{2}_{x^{2}}}\right]\hat{j}

The Attempt at a Solution

This is what I got for part b:

\vec{E} = 2kQ\left[\frac{1}_{x^{2}}} - \frac{1}_{x^{2}}}\right]\hat{j} = 0

I don't know how to get the book's answer of

\frac{6a^{2}}_{x^{4}}}\hat{j}

 

[alphysicist]

Hi Monocles,

This is for my intro physics 2 class

Homework Statement

Consider the charges Q at (-a, 0), -2Q at (0, 0) and Q at (a, 0). Such a combination of charges, with zero net charge and with zero net dipole moment, is called an electric quadrupole. a. Find the electric field along the x acis, for x > a. b. Show that, for x >> a, the electric field varies by x^-4. Find the coefficient.

Homework Equations

\vec{E} = \frac{kQ}_{r^{2}}\hat{r}

Here is the answer I got for part a which was correct.

\vec{E} = kQ\left[\frac{1}_{(x+a)^{2}}} + \frac{1}_{(x-a)^{2}}} - \frac{2}_{x^{2}}}\right]\hat{j}

The Attempt at a Solution

This is what I got for part b:

\vec{E} = 2kQ\left[\frac{1}_{x^{2}}} - \frac{1}_{x^{2}}}\right]\hat{j} = 0

I don't know how to get the book's answer of

\frac{6a^{2}}_{x^{4}}}\hat{j}

You do not want to set:

<br /> \frac{1}{(x+a)^2} \longrightarrow \frac{1}{x^2}<br />
because you'll lose the part you're looking for.

Have you seen the binomial series approximation formula for (1+y)^p when y is small? You can write your quantity \frac{1}{(x+a)^2} (and the other term, too) in that form and then use the formula to get the answer.