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 :)

Thanks a lot in advance for your answers!


Julien.

 

[BvU]

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 !

 

[JulienB]

@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 F_c 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.

 

[haruspex]

@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 F_c 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.

 

[BvU]

Spot on ! Well done .

Haru:

I just started electromagnetism

So the size of the spheres is ignored.

 

[JulienB]

@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]

@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"?