# Homework Help: Maximum electric force felt by the raindrop

1. Sep 29, 2011

### 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 29, 2011

### vela

Staff Emeritus
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)|.

3. Sep 29, 2011

### 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

4. Sep 29, 2011

### vela

Staff Emeritus
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.