# Electric Dipole as x l, how do I simplify this correctly?

1. Feb 26, 2013

### Yapper

Electric Dipole as x >> l, how do I simplify this correctly???

1. The problem statement, all variables and given/known data
ELECTRIC DIPOLE : 15.0 POINTS

A charge −Q is located at x=−l/2 and a charge +Q is located at x=l/2. Thus the separation between the two charges is l.

The total electric field on the x axis can be written as
E⃗ (x)=E(x)dˆ.

(a) What is the direction dˆ of the total electric field at any point on the x axis where x>l/2?

x^ (correct)

(b)What is E(x) as a function of x for x>l/2? Express your answer in terms of, if necessary, Q, l, x and the constant ϵ0 (if needed, enter pi for π, epsilon_0 for ϵ0).

(Q⋅l)/(2⋅π⋅ε⋅x^3⋅(1−(l/(2⋅x))^2)^2) (correct)

(c) Consider now the limit where x≫l, so that

(1±l2x)^−2≃1∓l/x

Express, in this limit, E(x) in terms of, if necessary, p, x and ϵ0. The quantity p=Ql is called the dipole moment.

p/(2⋅π⋅ε⋅x^3) (correct) how?????

2. Relevant equations

I already solved A and B with the equations. I really just need to know how to do C. Simplifying from (Q⋅l)/(2⋅π⋅ε⋅x^3⋅(1−(l/(2⋅x))^2)^2) to p/(2⋅π⋅ε⋅x^3) as x >> l.

3. The attempt at a solution

I substitute Q*l for P but I get stuck at the (1−(l/(2⋅x))^2)^2) part if I just have x = l then to me it should simplify to 9/16 but that's not it... then I try to use the (1±l2x)^−2≃1∓l/x approximation and i end up getting p (1 - l^2/x^2) / (2⋅π⋅ε⋅x^3) which when x >> l equals 0 i have no idea about how to make (1−(l/(2⋅x))^2)^2) go to 1 when x >> l, please help

2. Feb 27, 2013

### Yapper

Re: Electric Dipole as x >> l, how do I simplify this correctly???

A little typo correction,
(c) Consider now the limit where x≫l, so that

(1±(l/2x))^−2≃1∓l/x

Edit: I figured it out please ignore this post...

Last edited: Feb 27, 2013