**Will medal** coulomb's law
On number 2 I used the Pythagorean theorem to find the new distance, which i got .087m, found the force of -27.25N but it says its wrong? what am i doing wrong

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- rsst123

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- anonymous

Perhaps with the trigonometry function you messed up. Strength of electric field between 3 electrons are often presended with iscoleces triangle format but this one looks like an exception. Give me the precise details as to how you went about it so i can identify the problem.

- anonymous

Give me an update as soon as you are done showing the steps

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- rsst123

i converted the distances to meters, then i used the Pythagorean theorem, sqrt(.021^2+.085^2)=r, then i just plugged in the values forcoulombs law and got

- rsst123

-27.25N

- rsst123

- rsst123

i used F=kq1q2/r^2

- rsst123

i think im making a careless mistake but dont know where

- anonymous

k being the constant and r beingi the radius and yes that seems valid

- anonymous

let me investigate further

- rsst123

ill try doing it again too

- anonymous

If the r you used came from the hypotenuse, then you calculated the total force. They only want the horizontal part. You have to resolve the vector into components

- anonymous

Are you sure you used different charges to the last point of precision?

- rsst123

how would i find the x component? would i have to know the angle?

- anonymous

Yes as it appear you only calculated the electric field acting between the two particles, and not the one acting at 2.1cm away from the other. You would first calcualte the hypotenuse to account for the electric force acting between the two and then and the base of the triangle using trigonometry and that would be your answer. Make sure to use the length of the hypotenuse as the original electricl field from which to compute the one acting 2.1 cm away from the first particle.

- anonymous

Make sure to identify the length of the hypotenuse and then use the equation to find the electrical force acting between them, and then from that calculated eletrical force you will use trigonometry to calculate the corresponding force acting on the second longest side of the triangle or base of the triangle. The key is to shoot the formula once you get the length of the hypotenuse in meters and THEN use the trigonometry again.

- rsst123

figured it out, i found the angle from the x and y components by using tan and then used trig to find the force of the x component
Thank You!

- rsst123

the answer is -26

- anonymous

Great!!! My pleasure;)

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