# Homework Help: Finding net electric field at different locations

1. Sep 16, 2008

### nckaytee

1. The problem statement, all variables and given/known data
Two charges are placed on the x axis. One charge (q1 = +16.5 µC) is at x1 = +3.0 cm and the other (q2 = -19 µC) is at x2 = +9.0 cm. Find the net electric field (magnitude and direction) at the following locations.

(a) x = 0 cm
magnitude= ? N/C
direction= ?

(b) x = +6.0 cm
magnitude= ? N/C
direction= ?

I set E1=E2

[K(+16.5*10^-6)]/X^2 = [K(-19*10^-6)]/(X+.06)^2

After canceling everything out I get 16.5(X+.06)^2 = 19X^2

Then solving for X I get .8219 and -.0289

I am not sure what I am doing wrong. The (10^-6) should cancel out right?

2. Sep 16, 2008

### LowlyPion

First of all, welcome to PF.

But as to setting E1=E2 ... Why?

The E field at any point is

$$\sum \vec{E}$$

Perhaps a better approach is determining the E field contribution from each charge and adding them together remembering of course that you are adding E field vectors?

3. Sep 16, 2008

### nckaytee

So, It would be "[K(+16.5*10^-6)]/X^2 + [K(-19*10^-6)]/(X+.06)^2" instead?

I was following another example I found and now I am very confused.

4. Sep 16, 2008

### LowlyPion

Perhaps it would be less confusing if you dropped X from your equations. And then calculated each separately and added?

What is crucial is the distance from the charge. That and Coulomb's constant and the charge determine the E field contribution from that point.

So in the first one the E-field for the first charge at x= .03 m (dimensions in meters for the constant to be valid) resolves as:

$$\vec{E_1} = \frac{k*q_1}{r^2} = \frac{k*16*10^{-6}}{.009}$$

Then calculate the contribution for the negative charge that is at .09m.

Those two together determine the field at x=0.

$$\vec{E} = \vec{E_1} + \vec{E_2}$$

5. Sep 16, 2008

### nckaytee

When I do that I get -5111111.111

E= [(9x10^9)(16x10^-6)]/(.009) + [(9x10^9)(-19x10^-6)]/(.0081)

What am I doing wrong?

6. Sep 16, 2008

### LowlyPion

3 cm2 is .0009 not .009 as I originally typed. My mistake.

that makes it 16*107 - 2.1*107 = 13.9*107

7. Sep 16, 2008

### nckaytee

So, for part B I would just change the radius value?

8. Sep 16, 2008

### LowlyPion

Pretty much that looks like it.

Units are N/C - Newtons/Coulomb

9. Sep 16, 2008

### nckaytee

Thank you so much. I have a hard time understanding Physics and this really helped.

10. Sep 16, 2008

### LowlyPion

Great. Good Luck then.