What Is the Correct Value of q in Coulomb's Law Problem?

[jenduncan]

The net force on a 1nC charge is 0. What is q?

There are 3 forces. q is 2cm in the y direction away from a charge of 1nC. Another charge of 2nC is 4cm below q and 3cm to the right. The final charge is 4cm below q and 3cm to the left. Using Coulomb's law, I found the force of the two bottom charges to be 9.79e-6i+9.79e-6j and -9.79e-6i+9.79e-6j respectively. The final charge I found to be -22500q in the j direction. By adding all the forces together I find q to be 0.870nC. The answer in the back of the book is 0.68nC. Where have I gone wrong?

My final equation to solve for q was 1.96e-5 = 22500q

 

[whozum]

Force from first charge + Force fromsecond charge + Force from q = Force on 1nC charge = 0.

I think you missed this part which is why your off.

 

[quicknote]

+ 9.79e-6 + 9.79e-6 - 9.79e-6 = 22500q + 1.96e-5.

It seems that you have made a mistake in your final equation. The correct equation should be 1.96e-5 = 22500q + 9.79e-6 + 9.79e-6 - 9.79e-6, as the force from the final charge should be subtracted, not added. This would give a value for q of 0.68nC, which matches the answer in the book. It's important to carefully check all the signs and directions when using Coulomb's law to ensure accurate calculations. Additionally, make sure to use consistent units throughout the equation, as this can also lead to incorrect results.