Extrema of electrical potential

[FS98]

Homework Statement

A charge of 2 C is located at the origin. Two charges of −1 C each are located at the points (1, 1, 0) and (−1, 1, 0). If the potential φ is taken to be zero at infinity (as usual), then it is easy to see that φ is also zero at the point (0, 1, 0). It follows that somewhere on the y-axis beyond (0, 1, 0) the function φ (0, y, 0) must have a minimum or a maximum. At that point the electric field E must be zero. Why? Locate the point, at least approximately.

Homework Equations

φ = kq/r

The Attempt at a Solution

φ = 2k/y - 2k/sqrt(2-y)

The sqrt(2-y) comes from finding the distance between a given point on the y-axis and a particle through the Pythagorean theorom. The -2 coming from the fact that there are two negative charges with a magnitude of 1.

Then I took the derivative

dφ/dy = 2k(1/y^2-(1/2(2-y)^(3/2))

Set that equal to 0 and do some rearranging and crossing out.

y^2 = -2(2-y)^(3/2)

If I’ve done everything right up to this point, I don’t know how to solve for y in the final step.

 

[TSny]

The sqrt(2-y) comes from finding the distance between a given point on the y-axis and a particle through the Pythagorean theorom.

Note that sqrt(2-y) = 0 when y = 2. Does that seem right?

 

[FS98]

Note that sqrt(2-y) = 0 when y = 2. Does that seem right?

I did the trig over and got sqrt(y^2-2y+2) for the distance between a given value of y and the -1C particles.

After doing the rearranging again I have

y^2 = -2(y^2-2y+2)

Which I’m still unsure how to solve.

 

[TSny]

I did the trig over and got sqrt(y^2-2y+2) for the distance between a given value of y and the -1C particles.

OK. That looks right.

After doing the rearranging again I have

y^2 = -2(y^2-2y+2)

Which I’m still unsure how to solve.

I don't understand how you got this equation.

 

[FS98]

OK. That looks right.

I don't understand how you got this equation.

I think I differentiated incorrectly. I have

2k(1/y-(1/sqrt(y^2-2y+2)) for the potential energy.

I think the derivative is then
2k((2*y-2)/(2*(y^2-2*y+2)^(3/2))-1/y^2)

I’m not sure where to go from there.

 

[TSny]

I think I differentiated incorrectly. I have

2k(1/y-(1/sqrt(y^2-2y+2)) for the potential energy.

I think the derivative is then
2k((2*y-2)/(2*(y^2-2*y+2)^(3/2))-1/y^2)

OK. I think that's right.

I’m not sure where to go from there.

As you noted in your first post, you should set this equal to zero and solve for y. I believe you will only be able to get an approximate solution by using some numerical technique or by using a graph.