Calculate the leading behavior of the electrostatic potential $V(x)$ at large distances $|x| ≫ a$ for the following charge distributions:
a) One charge $q$ at the point $x_0 = ae_x$ and one charge $q$ at the point $−x_0$.
b) One charge $q$ at the point $x_0 = ae_x$ and one charge $−q$ at the point $−x_0$.
c) Two charges $q$ at the points $x_1 = ae_x + ae_y$ and $x_2 = ae_x − ae_y$ and two charges $−q$ at the points $−x_1$ and $−x_2$.
There are a few more questions but just very similar, the end of the day it's about vector calculus.
for a) the monopole term vanishes:
$$V_{dip}(r)=\frac{1}{4\pi \epsilon_0}\frac{P \cdot \hat{r} }{r^2}$$
So obviously $P=qd=q(2|x_0|)$, i guess that's the only simplification i can make
so $$V(r)=\frac{1}{4\pi \epsilon_0}\frac{q(2|x_0| \hat{x}) \cdot \hat{r} }{r^2}=$\frac{1}{4\pi \epsilon_0}\frac{qdcos \theta}{r^2}$$
for b) the monopole term dominates:
$$V_{Mon}=\frac{1}{4\pi \epsilon_0}\frac{2q}{r}$$
For c) Monopole term vanishes
$$P=aq(e_x+e_y)+aq(e_x-e_y)-aq(-e_x+e_y)-aq(-e_x-e_y)=4aqe_x$$
$$V(r)=\frac{1}{4\pi \epsilon_0}\frac{4aq \hat{x} \cdot \hat{r}}{r^2}=\frac{1}{4\pi \epsilon_0}\frac{4aq cos \theta}{r^2}$$
Is my understanding and calculation correct?