The question is,

A solid spherical conductor has a conical hole made at one end, ending in a point B and a small conical projection of same shape and size at the opposite end, ending in point A. If, now, a positive charge Q is transferred to the sphere then where the surface charge density will be greater at the point A or B?

My Approach: Surface charge density is inversely proportional to radius of curvature. But the question says that the conical hole and projection are of same size. So I think surface charge density should be same in both points but the answer sheet says Surface charge density at the point B would be greater than that of the point A. I don't understand why. Kindly explain.

## Best Answer

You have a conductor and in electrostatics it is an equipotential which means that when you bring a positive charge from infinity to any point on the conductor the same amount of work needs to be done.

If you have just a conducting sphere then from symmetry you can say that there must be a uniform surface charge density all over the sphere.

Now consider the protruding point $B$ with the same charge density as the rest of the sphere.

Then less work will have to be done to bring the positive charge to the point because the influence of the rest of the charged sphere will be less because those charges will be further away.

So to make the work done the same the surface charge density on the tip must be larger and as you quite rightly pointed out the surface charge density is related to the radius of curvature of the surface.

Now when the positive charge enters a hollow (with radius of curvature in the "opposite direction" to the point) the surrounding charges are now closer and so the surface charge density has to be lower to ensure that the work done in moving the positive from infinity the same.