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Question

The force of repulsion between two point charges is F, when they are d distance apart if the point charges are replaced by conducting spheres each of radius R and charges remain same the separation between the centre of the sphere d then the force of repulsion between them is

Equal to f

Less than f

Greater than f

Equal to f

Less than f

Greater than f

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Solution

In the present case, each sphere is not completely isolated from the environment because they are in the presence of each other. As a consequence they will interact. The fact that the two conducting spheres are charged implies that they will induce agregation of charges in certain regions of each other surfaces. To see this, think of both as spheres being negatively charged (and, therefore, will repel each other as required by the problem). These charges are free to move over the surface of their respetive spheres because we are dealing with a conductor. We can thus predict that the negative charge of one sphere will repel those of the other and tend to move and acumulate on the side of its respective sphere which is the furthest alway from the negative charges of the other sphere. Thus, if you have one sphere standing on the right and another one of the left, the negative charges of the sphere on the right will acumulate on the rightmost side of that sphere, and by symmetry, on the leftmost side of the sphere on the left. Now, we know that the amount of charge on both spheres remains the same during this process because charge is conserved. However, these charges will be further appart from each other now (when compared to the situation of point charges sitting at the geometrical center of the spheres) - and because Coulomb's Force decreases with the square of the distance, the force has to be smaller now.

Answer:- Less than F

Answer:- Less than F

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