Maximum electric force felt by the raindrop

[veevee]

Suppose 2 drops with equal charge q are on the x-axis at x=+-a. Find the maximum electric force felt by a third drop with charge Q at an arbitrary point on the y axis.

from the symmetry we can deduce that the x components cancel and the net force is in the y direction. the y component is same for both charges.

I used Coulomb's Law to get the equation

F=2((kqQ)/a²+y²)(y/sqrt(a²+y²) j(hat)

=(2kqQy)/(a²+y²)^3/2 j(hat)

I know that the next step is to find the y value for which the force is maximized, however I'm not too sure on how to do that. I thought maybe the y is max when the charges a are at the origin, but I don't think that the charges could be moved over to the origin for this calculation.

any suggestions?

 

[vela]

You're right. You can't move the charges on the x-axis; or mathematically speaking, a is a constant. What you can do is move the third charge, that is, vary y, to maximize |F(y)|.

 

[veevee]

hmm...

so since the length from the q charge to the Q charge is inversely proportional to the force felt by Q (1/y), the closer Q is to both q's, the more force it experiences. But if it was the lined up with them on the x axis, the net force would be 0. So, do I need to make it as close to the origin as possible? And would other equations, such as F=qE need to be incorporated to solve for y?

thanks for helping me

 

[vela]

As close to the origin as possible would be at the origin, where, you have already noted, the force is 0. If Q is very far away, again the force is essentially 0. So the maxima occur somewhere in between.

You already have the expression for the force. Just find where it's attains a maximum. This is a math problem now.