# Electric field of charged particles problem

1. Apr 2, 2009

### kevinf

1. The problem statement, all variables and given/known data

Two 9.00 µC charged particles are located on the x axis. One is at x = 1.00 m, and the other is at x = -1.00 m.

(a) Determine the electric field on the y axis at y = 0.900 m.

(b) Calculate the electric force on a -3.00 µC charge placed on the y axis at y = 0.900 m

the professor didn't really go over this type of question yet but i would like to get ahead and finish it. i know to apply the electric field equation (K)(q)/r^2 but i am not very good with working with components. can somebody explain this to me? thanks

2. Apr 2, 2009

### rl.bhat

The magnitude of the fields due to two charges will be same. But one field will go away from the charge and another towards the charge. Take vertical and horizontal components of the fields.
Vertical component = Esin(theta) and horizontal component = Ecos(theta) Find the resultant field.

3. Apr 2, 2009

### kevinf

i am not sure what you mean by one field will go away one will go towards the charge

4. Apr 2, 2009

### rl.bhat

One is repulsive and the other is attractive. Draw the diagram and see.

5. Apr 2, 2009

### kevinf

but i thought both charges are positive

6. Apr 2, 2009

### rl.bhat

Sorry. I didn't notice that. In that case both will be repulsive.
Vertical components will add up.And the horizontal components will cancel each other.

7. Apr 6, 2009

### kevinf

i did what you said but kept getting it wrong. wouldn't i use tangent first to find the bottom angles of the triangle that it would form and then use pythagorean theorem to find out what the lenght of the sides of the triangle are? i think the horizontal component should be 0. but i am having trouble with the vertical component of the problem. so i would do (9)*(Ke)/(1.81) and then multiply that by sin(angle that i calculated earlier) right? i am not getting the right answer.

8. Apr 6, 2009

### rl.bhat

I am getting 0.1796N.
If this is correct, show the details of your calculations.