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Force on charged spherical bodies

Physics Asked by Vivek Pandey on May 23, 2021

I came across this question:

The force of repulsion between 2 point charges is F. If the point charges are replaced by 2 spheres of finite radii carrying the same charges as the particles, then will the force between the spheres be the same as the original force?

I think we should treat the spheres as particles only, because whenever we calculate the electric field of a symmetrically charged sphere for an external point, we can assume it to be a particle. But the answer in the book says the force will decrease. Where am I going wrong?

2 Answers

If the spheres are non-conductors with a spherically symmetrical distribution of charge, they will act like point charges. If the spheres are conductors, the charges will repel each other and move to the opposite sides of each sphere.

Answered by R.W. Bird on May 23, 2021

The question is a bit vague, since the material composing the spheres was not indicated. However, to answer the question you need to know the charge distribution on the surface fo the spheres. Assume that the spheres are centred where the initial point charges were located, and assume that, given the finite radius of the spheres, the charge contained on the surface will be able to redistribute up to some extent. The more conductive the material, the higher the migration of the charge on the surface, the lower will be the repulsive force experienced by the spheres.

Why? You know that the electrostatic force scales with a factor of $1/d^2$, where $d$ is the distance between the (point) charges. Now, if the charge can move inside the material composing the spheres, since both spheres have the same charge (the force is repulsive), the charges will repel one another, moving at the furthest point on each respective sphere. Since the distance between repelling charges is higher, the repulsive force will consequently be lower.

I hope this helped.

Answered by magaDog on May 23, 2021

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