Hi there!
Let's say we have a point 'p' placed at a distance 'r' away from the origin, where r > > d.
The electric field from the +q charge will point towards the top-right, while the electric field from the negative chart will point towards the bottom-right.
Since both charges are of the same magnitude, the y-components will cancel out. We must solve for the x-component of the electric field.
We can begin by deriving an equation for the electric field.
[tex]E= \frac{kq}{R^2}[/tex]
[tex]R = \sqrt{(\frac{d}{2})^2 + r^2}[/tex]
We are solving for E in the x-direction, so:
[tex]E_x = \frac{kq}{R^2}sin\phi[/tex]
Substitute in the above, and:
[tex]sin\phi = \frac{\frac{d}{2}}{\sqrt{(\frac{d}{2})^2 + r^2}} = \frac{d}{2\sqrt{\frac{d^2}{4} + r^2}}[/tex]
Calculate Ex for one charge:
[tex]E_x= \frac{kq}{\frac{d^2}{4} + r^2} * \frac{d}{2\sqrt{\frac{d^2}{4} + r^2}}[/tex]
Simplify:
[tex]E_x = \frac{kqd}{2(\frac{d^2}{4} + r^2)^{3/2}}[/tex]
There are two charges, so:
[tex]2E_x = E_x = 2(\frac{kqd}{2(\frac{d^2}{4} + r^2)^{3/2}}) = \frac{kqd}{(\frac{d^2}{4} + r^2)^{3/2}}[/tex]
To find the field if r > > d, we can begin by factoring out r² from inside the parenthesis:
[tex]E_x = \frac{kqd}{(\frac{d^2}{4} + r^2)^{3/2}} \\
\\
=\frac{kqd}{r^3(\frac{d^2}{4r^2} + 1)^{3/2}}[/tex]
Terms with d/r go to 0, so:
[tex]\frac{d}{r^3} = 0 \\
\\
\frac{d^2}{4r^2} = 0 [/tex]
So:
[tex]
E_x =\frac{kq(0)}{(0 + 1)^{3/2}} = \boxed{0 \frac{N}{C}}[/tex]
**We can also think of this situation as d ≈ 0. As the 'r' increases and becomes MUCH greater than 'd', the charges appear to be right next to one another (d ≈ 0). If we plug in d = 0 into our equation:
[tex]
E_x =\frac{kq(0)}{(\frac{0^2}{4} + r^2)^{3/2}} = 0 \frac{N}{C}[/tex]
