Show simplification of dipole's electric field for certain case

[s3a]

Homework Statement

I'm just trying to show that the equation in the attached image becomes E = Q/(2π $ϵ_0$ $r^2$) * $i^$, where $i^$ is i-hat, the unit vector for when r is the smallest that it could be, which is when r = d/2.

I'm doing this with reference to problem 1.2 (and I'm including problem 1.1 so that you have the background information for problem 1.2).

Homework Equations

The equation in the image as well as E = Q/(2π $ϵ_0$ $r^2$) * $i^$.

The Attempt at a Solution

My attempt is attached as MyAttempt.pdf.

Is it a coincidence that I'm getting double the value shown in problem 1.2?

Any help in getting to show that the vector equation for the electric field of a dipole breaks down to E = Q/(2π $ϵ_0$ $r^2$) * $i^$ would be greatly appreciated!

 

[Bashkir]

You are approaching this well, but is easier to why this result comes if you examine it like this.

The x-components of the electric field add, so we say that,

E_t = E_1 + E_2\\<br /> E_1 = \frac{1}{4\pi\epsilon_o}\frac{Q}{r^2}sin\theta

Which is the same as E_2 because both value of sines are positive. Even though we have a specific value for what r should be, to justify our solution we will define r using the Pythagorean theorem. So the total electric field then becomes.

E_t = \frac{2}{4\pi\epsilon_o}\frac{Q}{y^2 + \frac{d}{2}}sin\theta\\<br /> sin\theta = \frac{d}{2}{r}\\<br /> E_t =\frac{1}{4\pi\epsilon_o}\frac{Qd}{r^3}\\<br /> E_t = \frac{Qd}{4\pi\epsilon_o}\frac{8}{d^3}\\<br /> E_t = \frac{Q}{2\pi\epsilon_od^2}

Then remember that your d value is actually
d^2 = \frac{d^2}{4} = r^2

This should give you the answer that you are searching for.

 

[s3a]

Thanks for your answer.

Isn't what I got in the PDF of my work (in my opening/first post of this thread) $E_t$?

It seems that you made a few typos, so in an attempt to overlook the typos, I think your work results in $E_t$ = Q/(2π $ϵ_0$ $r^2$), such that $E_1$ = $E_2$ = Q/(4π $ϵ_0$ $r^2$).

I'm mentioning this because my $E_t$ seems to be double yours (which in turn is double $E_1$ and $E_2$, respectively – such that my $E_t$ is four times $E_1$ and $E_2$, respectively), so while I think I see what you're trying to say, I'm still not getting the algebra to confirm it.

Could you please elaborate on this situation?