We have proton and electron and distance between them is $d$. If we choose $z$ axis along this dipole, net electric potential created by dipole on the point A is equal to summation of the electric potential created by electron and proton. If we defined the distance between point A and electron as $r_1$ and distance between point A and proton is defined by $r_2$. Then, we choose a middle point of the dipole and distance between this point and A is $r$, and angle between $z$ axis and $r$ is called $\alpha$. When we are in this step, we can approximate for $r_2$ minus $r_1$. I could not understand this approximation. How we can defined as below? How it is become?

$$ r_2-r_1 \approx d\cos \alpha$$

If my question is not clear please inform me.

  • $\begingroup$ I could not see because there is no right triangle but maybe we assume this triangle as an right triangle and we get this approximation, I am not sure, too. $\endgroup$ – elif cetiner Mar 23 '16 at 23:09

I tried to depict the general idea behind the approximation. I hope it makes sense. In essence, it relies on the segment from $P$ to $+q$ and from $P$ to the perpendicular line intersecting $r_2$ being about equal, so the difference is the labeled leg of the triangle shown in the upper corner of the diagram, where the angle $\alpha$ is adjacent to the side $r_2-r_1$, and has a hypotenuse of $d$, so using some simple trig, that side is also $d \cdot \cos{\alpha}$.

Dipole geometry

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