Calculating Electric Field of Two Charged Particles on the y-axis

[Carrie]

Homework Statement

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

Homework 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. [/B]

 

[SammyS]

Homework Statement

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

Homework 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. [/B]

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))​

 

[Carrie]

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

 

[SammyS]

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

Are you rounding off any intermediate results?

 

[Carrie]

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

 

[SammyS]

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

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)).

 

[Carrie]

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)).

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?

 

[SammyS]

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?

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

 

[Carrie]

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