# Two Point Charges Net Electric Field

1. Feb 1, 2010

### Octoshark

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

Two point charges lie on the x axis. A charge of -2.5 µC is at the origin, and a charge of +8.5 µC is at x = 10.0 cm.

What is the net electrical field at points x = +4 and at x = -4?

2. Relevant equations

E = kq/r^2

3. The attempt at a solution

For x = +4...

E= 9x10^9(8.5x10^-6)/.06^2 = 21250000
E= 9x10^9(-2.5x10^-6)/.04^2 = -14062500

Net E = 7187500 (which is wrong)

How do I solve this? Thanks.

2. Feb 1, 2010

### theRIAA

use absolute values for charges and then add/subtract them up qualitatively.

the -2.5uC will make the E field LARGER at point x=+4 so you would add it instead of subtract.

3. Feb 1, 2010

### Octoshark

That didn't work.

4. Feb 1, 2010

### theRIAA

I'm pretty sure it should work. What's the correct answer for both positions?

5. Feb 1, 2010

### Octoshark

It doesn't give me the right answer until I submit the right answer. I entered 35312500 for x = +4 and it says that is incorrect. (I submit the answers online)

6. Feb 1, 2010

### theRIAA

the answer is 3.5*10^7 after sig figs.

7. Feb 1, 2010

### Octoshark

It says that is wrong too :yuck: