# Calculating Electric Field

1. Feb 17, 2016

### Carrie

1. The problem statement, all variables and given/known data
Two 1.00µC charged particles are located on the x axis. One is at x = 1.00 m, and the other is at x = -1.00 m.
Determine the electric field on the y axis at y = 0.400 m

2. Relevant equations
Electric Field = ke * q / r^2

3. The attempt at a solution

I know the x-components cancel out, so we're only left with the y-component.
E= 8.99*10^9 * (1*10^-6 C / (square root(1^2 + .400^2)) and then you have to take the y-component into account, so you multiply that by sin(.400/(square root(1^2 + .400^2))... and then multiply everything by 2 because there are two different charged particles. I'm getting an answer of 5625 N/C, but apparently the answer is 5760 N/C.

2. Feb 17, 2016

### SammyS

Staff Emeritus
According to the following, you only divided by r, not r2 .

E= 8.99*10^9 * (1*10^-6 C / (square root(1^2 + .400^2))

3. Feb 17, 2016

### Carrie

When I put it in my calculator, I did square it... oops, just forgot to type it here. Still getting 5625.

4. Feb 17, 2016

### SammyS

Staff Emeritus
Are you rounding off any intermediate results?

5. Feb 17, 2016

### Carrie

Nope, I put it in exactly how I did here without rounding off the calculation of the hypotenuse.

6. Feb 17, 2016

### SammyS

Staff Emeritus
Maybe this is the problem. The following

.400/(square root(1^2 + .400^2))

is the sine of the angle. Don't take the sine of this. That would be sin(sin(θ)).

7. Feb 17, 2016

### Carrie

What do you mean? I thought that since sine is OPP/HYP and the opposite would be 0.400 and the hypotenuse would use Pythagorean theorem and it would be square root(1^2 + .400^2, so it would be sin(.400/(square root(1^2 + .400^2))... right?

8. Feb 17, 2016

### SammyS

Staff Emeritus
Yes. That, 0.400/square root(1^2 + .400^2), IS the sine. Don't take the sine of the sine.

9. Feb 17, 2016

### Carrie

Ohhhh wow, that was it! Thank you so much! I didn't even realize.