How Do You Calculate Electrostatic Force Between Charges?

[BoldKnight399]

Two positive point charges, each of which
has a charge of 1.5 × 10−9 C, are located at
y = +0.50 m and y = −0.50 m.
The Coulomb constant is 8.98755 ×
10^9 N · m2/C2.
a) Find the magnitude of the resultant elec-
trical force on a charge of 1.0×10−9 C located
at x = 0.55 m.
Answer in units of N.

So I drew out a force diagram and thought that the equation would be:
Fel=kc(Q1*q/d1^2)+kc(Q2*q/d2^2) where the q is the 1.0e-9C charge.
Fel=(8.98755e9)(1.5e-9*1.0e-9/.0025)+(8.98755e9)(1.5e-9*1.0e-9/1.1025)
so Fel=5.404757e-6

Apparently that was the wrong answer. Does anybody know or want to explain how I should now re-examine the problem?

 

[Doc Al]

The force from each charge is a vector. You have to add them like vectors, not numbers--the direction matters. Hint: What happens to the y-components?

Also: What's the distance between the charges?

 

[BoldKnight399]

yeah...I missed the fact that q was on the x axis. makes sense now. Thank you for your help!

 

[rajeshpamula]

I have a question about the operation of the electrostatic force. I solved the Maxwell's equations in a pure static electric conditions and the result I got does not look like a wave equation. If that is the case, then what is the speed of electrostatic interaction in free space? And what is the messenger particle?

 

[rdl]

Your approach to the problem is correct, but there may be a mistake in your calculation. The correct answer should be:

Fel = (8.98755e9)(1.5e-9*1.0e-9/(0.55^2+0.5^2)) = 4.9431e-6 N

This is because the distance between the charge at x = 0.55 m and the charge at y = +0.50 m is not 0.0025 m but the hypotenuse of a right triangle with sides 0.55 m and 0.50 m.

Also, make sure to use the correct units for the distance, which should be in meters. So it should be:

Fel = (8.98755e9)(1.5e-9*1.0e-9/(0.55^2+0.5^2)(0.01^2)) = 4.9431e-6 N

I hope this helps!