Binomial Theorem and Electric Field question

[doubleUTF]

Homework Statement

I'm having a lot of trouble solving for part b as I am unable to correctly apply the binomial theorem to this approximation. The problem is shown below:

Three point charges are distributed: a positive charge +2Q in the center, and a pair of negative charges -Q, a distance a to its left and right.

You want to find the electric field E at point P, a distance r to the right of the positive charge.

a) Write an exact expression for the x-component of E at point P (in terms of Q,r,a, etc.

b) Using the binomial theorem, write a simpler approximation to the expression you gave in part b, which is valid when r>>a

Homework Equations

Binomial theorem: (1+x)^n=1+nx+n(n-1)x^2/2!+...

The Attempt at a Solution

a) I added the vector sum of all forces and got 1/(4pi epsilon knot)(-Q/(r+a)^2+(2Q)/r^2+(Q)/(r-a)^2

b) The professor gave out the answer for this part which is 6kQa^2/r^4, but we need to show how we got this. I go as far as KQ(2-(a-2a/r)-(1+2a/r))r^2, but I don't know how to proceed further. I wish I knew how to type out the work more neatly so you can see clearly. I would greatly appreciate any feedback.

 

[Fredrik]

I don't see how the binomial theorem is relevant here. What you need to know is the sum of a geometric series:

$$1+z+z^2+z^3+\dots=\frac{1}{1-z}$$

This holds for for all complex z with |z|<1. Proof:

$$1+z+z^2+\dots+z^n=\frac{(1-z)(1+z+z^2+\dots+z^n)}{1-z}=\frac{1-z^{n+1}}{1-z}$$

Now take the limit $n\rightarrow\infty$.

To solve b), I suggest you take -Q/r^2 outside the parentheses. Use the formula for the sum of a geometric series with z=a/r and z=-a/r. Keep terms up to second order and ignore higher order terms.

 

[doubleUTF]

Not to blow off your response, but our professor specifically asked that we utilized the binomial theorem to write a simpler approximation. Plus we were never taught to apply geometric series to this problem. This is what I did:$$K = \frac{1}{{4 \pi \varepsilon_0}$$

from part (a), $$(E_{net})_x = KQ (\frac{2}{r^2} - \frac{1}{(r+a)^2} -\frac{1}{(r-a)^2}) \vec{i}$$

and $$\frac{1}{(r-a)^{2}} = (r-a)^{-2} = r^-2[1-\frac{a}{r}]^-2$$
& $$\frac{1}{(r+a)^2} = r^-2 [1+\frac{a}{r}]^-2$$

So when r>>a, $$(E_{net})_x=KQ (\frac{2}{r^2} - \frac{1}{(r+a)^2} - \frac{1}{(r-a)^2}) \vec{i}=\? = \frac{6KQa^2}{r^4}$$
Sorry I'm having trouble with this LaTeX code, but hopefully you can make out what I did.
I'm lost in the part where I placed the question marks.

 

[chaoseverlasting]

OK, its not that bad. In your expression for the electric field, $$kq(-1/r^2+ (r+a)^{-2} -(r+2a)^{-2}, ((r+1)-1)^{-2}$$. Here, expand $$(r+a)^{-2}, (r+2a)^{-2}$$ binomially, giving you $$1-2ra +(higher terms) +1-ra +higher terms+1-2(r-1)$$. Ignoring the higher terms of the expansion as r<<a, you get $$1-3ra-2r$$. Dunno how you got that answer though.

 

[Fredrik]

I find it very hard to believe that the binomial theorem can have any relevance here. You have already found that

$$E=-\frac{KQ}{r^2}\big(\big(\frac{1}{1-z}\big)^2-2+\big(\frac{1}{1+z}\big)^2\big)$$

where $z=a/r$. It's easy to simplify this to the answer you want if you just do what I suggested before.

Edit: I got the opposite sign in the final answer.