# Electric Field between 2 Point Charges

1. Jan 27, 2008

### Goldenwind

[SOLVED] Electric Field between 2 Point Charges

1. The problem statement, all variables and given/known data
http://session.masteringphysics.com/problemAsset/1011305/15/1011305A.jpg

Two point charges are placed on the x axis. The first charge, q1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis.

Find the electric field at the origin, point O.

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

2. Relevant equations
E = F/q
F = kq1q2/r^2
E = p / (2pi * e0 * x^3)
p = qr

q1 = 8*10^-9
q2 = 6*10^-9
r = 9+16 = 25 (The distance between the two point charges)
e0 = 8.85*10^-12
x = r (Assumably, not for certain)

3. The attempt at a solution
Since both charges are along the x axis, I conclude that they do not pose any influence on the y coordinate of the field, therefore the y coordinate is 0.

The x coordinate can be computed via E = F/q.
We are trying to calculate E.
F can be calculated via F = kq1q2/r^2
Issue is, what do we use for q?

Alternatively, we could use p = qr to find p (But again, which q to use?)
Then we can use E = p / (2pi * e0 * x^3)
(But then which x do we use? 9, 16, or 25? Distance between a point and (0,0), or distance between both points?)

2. Jan 27, 2008

### foxjwill

I think you're making this too hard. All you need to do is find the electric field at the origin due to $$q_1$$ and add it to the electric field at the origin due to $$q_2$$. In other words,

$$\vec{E} = \vec{E}_1 + \vec{E}_2$$