# Force acting on a charge

## Homework Statement

A charge +2nC is located at (0, 2 m). Another +2 nC charge is
located at (0, -2 m). A third charge, +3 nC, is located at (2 m, 0).
Calculate the total force (in vector form) acting on charge +3nC
from both +2nC charges.

## Homework Equations

Coulomb's law in vector form

## The Attempt at a Solution

I worked out the force on q3 due to q1 to be:

9.54 nano Newton in the positive x direction and
-9.54 nano Newton in the negative y direction

Next, I worked out the force on q3 due to q2 and got:

9.54 nano Newton in the positive x direction and in the positive y direction.

Adding these together, the component would be zero and the total force would be 19.08 nano Newton in the x direction. This would be my answer. However, the answer given is 9.537 nano Newton in the x direction.

Any help is appreciated!
Thanks

Last edited:

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gneill
Mentor
What is the difference between "in the x direction" and "in the positive x direction"? What is the difference between "in the x direction" and "in the positive x direction"? Nothing, I've changed it to avoid any more confusion I worked out the force on q3 due to q1 to be:

9.54 nano Newton in the positive x direction and
-9.54 nano Newton in the negative y direction

So, a force in the positive y-direction?

19.08 nano Newton in the x direction. This would be my answer. However, the answer given is 19.08 nano Newton in the positive x direction.

Please correct and I'll see if I can help...

So, a force in the positive y-direction?

Please correct and I'll see if I can help...
Ahh sorry, the given answer is 9.537 nN.

gneill
Mentor
So it looks like you've mucked up a factor of two somewhere. Can you show your calculations? What value did you use for the distance in Coulomb's law?

So it looks like you've mucked up a factor of two somewhere. Can you show your calculations? What value did you use for the distance in Coulomb's law?
I used 4 (two squared) for the position vector and root 2/2 for the unit vector. This is for the x component for the force on q3 exerted by q1.

gneill
Mentor
Let's take the case of q1 and q3. q1 is at (0,2), while q3 is at (2,0). The distance betwixt is:

$$\sqrt{(2 - 0)^2 + (0 - 2)^2} = \sqrt{4 + 4} = 2 \sqrt{2}$$ (meters)

r13 = (2, -2), unit vector r = root 2 / 2 in the x direction - root 2 / 2 in the y direction,

F13 = Q1Q3 * unit vector / 4 PI Epsilon nought r^2

(with r = 2)

Ahh ok, now I see where I was going wrong!

Thanks for walking me through that gneill, massively appreciate it.