# Forces Of Charge In An Electric Field

1. May 17, 2009

### Air

1. The problem statement, all variables and given/known data
A charge of $$8\times 10^{-5}$$C is placed in an electric field $$(3000\b{i}-600\b{j})$$N/C. What are the (a) magnitude and (b) direction (relative to the positive direction of the $$x$$-axis) of the electrostatic force?

2. Relevant equations
$$F=Eq_0$$

3. The attempt at a solution
(a) $$F=(0.24\b{i}-0.048\b{j}) \implies |F| = 0.245$$N
(b) How is direction determined? I understand it's positive charge thus the field is always pointing away from the charge. By finding angle, $$\theta = \tan ^{-1}\left(-0.048/0.240) = -11.3^o$$, but what does the force direction actually point in? Away from origin, towards origin? The angle I found, I cannot relate to theory. Can someone explain the direction.

2. May 17, 2009

### FedEx

You have just to find the direction wrt x axis. And you have found it. There is nothing like towards or away the origin... Away or towards depend upon the position of the charge

3. May 17, 2009

### nickjer

You have found the angle with respect to the x-axis as FedEx has said. I also recommend drawing a picture of the electric field, and since the charge is positive, the force will be pointing in the same direction as the field. Just so you can understand the concept and not just see numbers.

4. May 17, 2009

### Air

It's a positive charge so field will be radially outwards. Is force also in those direction. If that is the case, then it will cancel out as there will be forces in opposite direction.

5. May 17, 2009

### nickjer

No, you don't include the field from the charge since it doesn't feel a force from itself. You only include the electric field that acts on the charge.

6. May 17, 2009

### Air

But there isn't a field which acts on it.

There's only it's electric field.

7. May 17, 2009

### nickjer

Yes, that is another way of putting it. I was just discussing the sketch you draw. You don't need to draw the field lines of the point charge since you don't take it into consideration when determining the force on it.

Edit: Did you edit your last post? Because what you said originally was right, but now what you said is wrong. HallsofIvy explains what is wrong below.

Last edited: May 17, 2009
8. May 17, 2009

### HallsofIvy

Staff Emeritus
No, the problem specifically said "A charge of 8 x 10-15C is placed in an electric field $$(3000\b{i}-600\b{j})$$ N/C". That is the field acting on it.

9. May 18, 2009

### Air

Oh yes. So the resolved direction would give that the field is South-East?