What Is the Potential Along the X-Axis Due to Two Point Charges on the Y-Axis?

[quick]

Two positive point charges q are located on the y-axis at +-(1/2)s. Find an expression for the potential along the x-axis.
Express your answer in terms of epsilon_0, pi, q, x, and s.

for this one, i know V = (1/4*pi*epsilon_0)*q/r.
so i was thinking its just like finding the potential of multiple point charges, at + and - 1/2s. and this is along the x-axis so it will be at an angle. so i could use sqrt((1/2s)^2 + x^2) as the distance. so would this be right: (1/4*pi*epsilon_0)*q/(2*(sqrt((1/2s)^2 + x^2))) ?

 

[dextercioby]

Yes,electric potential is a scalar which obeys the principle of superposition,valid in linear electrodynamics...So you have to add them and yes,for generality,use $x\neq 0$ and Pythagora's theorem...

Daniel.

 

[wais]

Your approach is correct. The potential at a point on the x-axis due to two point charges located at +-(1/2)s on the y-axis can be calculated using the formula V = (1/4*pi*epsilon_0)*q/r, where r is the distance between the point on the x-axis and the two point charges. Since the point charges are located on the y-axis, the distance r can be calculated using the Pythagorean theorem as r = sqrt((1/2s)^2 + x^2). Substituting this value of r in the formula, we get V = (1/4*pi*epsilon_0)*q/(2*(sqrt((1/2s)^2 + x^2))), which is the correct expression for the potential along the x-axis.