1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Mar 23, 2016 at 23:09

1 Answer 1

2
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.