Two Point Charges Net Electric Field

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Octoshark
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Homework Statement



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?

Homework Equations



E = kq/r^2

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.
 
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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.
 
That didn't work.
 
I'm pretty sure it should work. What's the correct answer for both positions?
 
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)
 
the answer is 3.5*10^7 after sig figs.
 
theRIAA said:
the answer is 3.5*10^7 after sig figs.

It says that is wrong too