# Electric Fields and charged particles

1. Dec 14, 2014

### TheExibo

Let's say there's a question where there are two charges placed, and a point in between and above the line connecting the two charges forms a triangle like this:

http://session.masteringphysics.com/problemAsset/1413940/2/p19.5.jpg

What would one have to do in order to find the electric field at point P? Would it be having to use the component method to find the x and y values, then using Pythagorean theorem and so on?

(I couldn't use the template for this post)

2. Dec 14, 2014

### Bystander

Don't doubt yourself when you do understand something. You're in business.

3. Dec 14, 2014

### TheExibo

I tried using that to find the y values, added them up, yet its too large. The net x value in the actual question is 0 in the center of the two charges.

4. Dec 14, 2014

### Bystander

Check.
Let's see your numbers for the y component.

5. Dec 14, 2014

### TheExibo

The height of the triangle is the square root of 3m. An electric field at this distance away from one of the points would have a magnitude of 17980N/C. Multiplying that by two because there are two charges gives a magnitude of 35960N/C upward. The correct answer is 1.2x10^5N/C.

6. Dec 14, 2014

### Bystander

It's an attractive short cut, and like all shortcuts, it's the farthest distance between two points. You have to sum each charge increment times the inverse square of its distance from the point. If the distance is much greater than the distance between charges, what you've done is a reasonable approximation, but not the case here.
Take one more swing at it, please.

7. Dec 14, 2014

### TheExibo

I'm not sure what you mean here. Is the distance in the y-axis? The hypotenuse? Also, I check and the answer in the book isn't correct. So the answer that I got is not correct?

8. Dec 14, 2014

### Bystander

Yes.
You want the absolute distance between each charge and the point at which you're calculating the field.
I was going to get to that.
More correct than the book, but you need to use the hypotenuse.
One more time.

9. Dec 14, 2014

### haruspex

Remembering that these are vectors to be summed.
That's way too much.

10. Dec 14, 2014

### TheExibo

The electric field intensity at a point 2m away from one of the charges is 13485N/C which is the hypotenuse. I know the angle at the bottom two vertices are 60 degrees. Doing (13485N/C)sin60 gives an answer of 11678N/C. Multiplying that by two gives 23357N/C upward. Is this the correct process?

11. Dec 14, 2014

### Bystander

You nailed it, round to the appropriate number of significant figures, and have a happy.

12. Dec 14, 2014

Thanks!! :D