Why is one of the solutions obtained using Coulomb's Law not valid?

[Saladsamurai]

[SOLVED] Question about Coulomb's Law

I just solved this a problem in my text: Particle 1 of charge +1.0 uC and particle 2 of charge -3.0 uC are held at a separation L=10.0 cm on an x-axis. If particle 3 of unknown charge q3 is to be placed such that the net electrostatic force on it from 1 and 2 is zero, what must its x and y coordinates be?

Okay, I solved this using Coulomb's Law which gave me a quadratic. Thus, I got two solutions. My question is: why is one of them not valid?

I got x=.0366 and x=-.14. -.14 m is the correct answer. Why does the .0366 work? And if it does not, why did the quadratic produce it?

It may be obvious, but I am just not seeing it now.

Thanks

 

[Dick]

It's obvious. A charge of either sign placed between two opposite charges will experience a force from both in the same direction. It has to be outside of the region to be in even unstable equilibrium. Quadratics often give an unphysical solution.

 

[Saladsamurai]

I knew it had to do with the signs. . . I just couldn't put my finger on it. Thanks Dick! Now I know to look for that.

 

[Mamun3680]

Hello !
Can anyone help me to solve the below problem on Coulomb's law please..
One Metal rod with -1000C charge is hanging on the earth. One another charged metal ball which mass 1 kg is tie by a rope & it floating 1m away from the rod. How much charge is containing by the ball??

 

[rwisz]

for your help!

I would like to first commend you for solving the problem using Coulomb's Law and obtaining two solutions. It shows that you have a good understanding of the concept. Now, to answer your question, one of the solutions may not be valid because it does not satisfy the given conditions of the problem. In this case, the condition is that the net electrostatic force on particle 3 from particles 1 and 2 must be zero. Therefore, the solution that does not satisfy this condition is not a physically possible solution. The reason the quadratic produced it is because it is a mathematical solution, but not all mathematical solutions have a physical significance. It is important to always check the solutions obtained and see if they satisfy the given conditions of the problem. I hope this helps clarify your doubt. Keep up the good work!