# Potential energy

## Homework Statement

Two point charges are arranged as shown on the x axis. q1 = +7.00 nC,
and q2 = −5.00 nC. a = 25.0 cm.
Calculate the position on the x axis, outside of the charges, where the
electric potential is zero

|......a........|
O----------O
q1..............q2

(ignore the periods)

## The Attempt at a Solution

V=V1+V1

kq1/d = -kq2/d

k(7nC)/d = -k(-5nC)/(d-25)

then solve.

I get the correct answer but my question is, why is it d-25 and why is it under q2? My first approach I had.

k(7nC)/(d+25) = -k(-5nC)/d

Last edited:

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It depends on what d is. your first approach used d as the distance from q2, and the secontd approach used d as the distance from q1. And in the equation kq1/d = -kq2/d you use both at the same time. This should hae been kq_1/d_1 = -kq_2/d_2

I would have used x for the position of the point with zero potential, the positions of the charges are x_1 = 0, and x_2 = 25
would have d_1 = |x - x_1| and d_2 = |x - x_2| (|x| is the absolute value of x)
if you look for a point to the right of q_2 you can simply have
d_1 = x - x_1 = x
d_2 = x - x_2 = x - 25