# Maximal repellence between charged spheres

• JulienB

## Homework Statement

Hi everybody! I just started electromagnetism, and I'd like to make sure I get the concepts correctly before going further. I got this homework:

A charge is shared between two spheres. How big must the share of charge relative to one another be, so that there is the maximal repellent force between them with a given distance r.

## Homework Equations

Coulomb force: $$F_c = \frac{1}{4 \cdot π \cdot ε_0} \cdot \frac{q_1 \cdot q_2}{r^2}$$

## The Attempt at a Solution

Well I know that the charges must have the same sign to make the coulomb force repellent, so my first guess would be that q1 and q2 will repel each other at most when there are equal, that is q1 = q2 = ½⋅∑q. Is that correct? Sounds a bit too easy, but it's only the first homework :)

Julien.

Good guess. Now you need to underpin it. Total charge Q is given, say a fraction f goes to q1. Express the force in terms of f and maximize !

@BvU Cool, thanks for your answer! Here's the coulomb's force between the two charges:

$$F_c = \frac{1}{4 \cdot π \cdot ε_0} \cdot \frac{a \cdot q \cdot (1 - a) \cdot q}{r^2} = \frac{1}{4 \cdot π \cdot ε_0} \cdot \frac{(a - a^2) \cdot q^2}{r^2}$$

That means Fc is maximal when (-a2 + a) is maximal, so I differentiate it and find:

$$\frac{d}{da} (-a^2 + a) = -2a + 1 \implies a = \frac{1}{2}$$

Therefore:

$$q_1 = q_2 = \frac{1}{2} \cdot q$$

What do you think?

Julien.

@BvU Cool, thanks for your answer! Here's the coulomb's force between the two charges:

$$F_c = \frac{1}{4 \cdot π \cdot ε_0} \cdot \frac{a \cdot q \cdot (1 - a) \cdot q}{r^2} = \frac{1}{4 \cdot π \cdot ε_0} \cdot \frac{(a - a^2) \cdot q^2}{r^2}$$

That means Fc is maximal when (-a2 + a) is maximal, so I differentiate it and find:

$$\frac{d}{da} (-a^2 + a) = -2a + 1 \implies a = \frac{1}{2}$$

Therefore:

$$q_1 = q_2 = \frac{1}{2} \cdot q$$

What do you think?

Julien.
Strictly speaking, that only shows it is a local extremum. A bit more work is needed to show it is a local maximum, and a bit more to show it is a global maximum.
An easier way avoids calculus. Get a(1-a) into the form (some constant)+/-(some function of a)2.

I note that it does not say whether the spheres conduct. If they do, your expression for the repulsive force is not correct. It becomes a lot more complicated, even to the point that two conducting spheres of like charge can attract.

Spot on ! Well done .

Haru:
I just started electromagnetism
So the size of the spheres is ignored.

@haruspex @BvU Thanks for your answers. The 2nd derivative being negative, it means that it is a local maximum. Since the first derivative is equal to zero for only one value, this local maximum must be a global maximum. Does that work?

@haruspex I don't really get your other method. Can you describe it a little more please?

Thanks a lot to both of you!

Julien.

@haruspex @BvU Thanks for your answers. The 2nd derivative being negative, it means that it is a local maximum. Since the first derivative is equal to zero for only one value, this local maximum must be a global maximum. Does that work?

@haruspex I don't really get your other method. Can you describe it a little more please?

Thanks a lot to both of you!

Julien.
Local versus global can also be complicated by the domain of the function. E.g. the maximum of 1/(1+x+y) in the first quadrant.

Did you try expressing a-a2 in the form I said? Do you know how to "complete the square"?