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

[tex]\vec{E} = \frac{kQ}_{r^{2}}\hat{r}[/tex]

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

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

The Attempt at a Solution

This is what I got for part b:

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

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

[tex] \frac{6a^{2}}_{x^{4}}}\hat{j} [/tex]

 

[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

[tex]\vec{E} = \frac{kQ}_{r^{2}}\hat{r}[/tex]

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

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

The Attempt at a Solution

This is what I got for part b:

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

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

[tex] \frac{6a^{2}}_{x^{4}}}\hat{j} [/tex]

You do not want to set:

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

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

 

[baba89]

First of all, great job on part a! Your solution is correct and shows a good understanding of the concept of an electric quadrupole.

For part b, it seems like you have made a small mistake in your calculation. Let's take a closer look at the electric field equation for an electric quadrupole:

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

When you substitute in x >> a, you get:

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

Notice that when you combine the fractions, you get:

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

This is why you got an answer of 0 in your attempt. However, the book's answer is actually:

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

To get the book's answer, you need to expand the fractions before combining them:

\vec{E} = 2kQ\left[\frac{1}{x^{2}} - \frac{1}{x^{2}}\right]\hat{j} = 2kQ\left[\frac{x^{2}}{x^{4}} - \frac{x^{2}}{x^{4}}\right]\hat{j} = \frac{6a^{2}}{x^{4}}\hat{j}

I hope this clears up any confusion and helps you understand the concept better. Keep up the good work!