# Induced charge on a conducting sphere sliced by a plane

• Hellec408
Hellec408
Thread moved from the technical forums to the schoolwork forums
We are given a conducting solid sphere, and it is cut by a plane which has a minimum height r/2 from the centre of the sphere, which has radius r.

A charge Q is given to the smaller part of the conductor, and it is required to find the induced charge on the curved and flat surfaces of the other hemisphere, and also the force of interaction between the two parts of the sphere.

My initial approach was to apply Gauss law, as we know that electric field is always perpendicular to the conductor surface, and hence we can claim that field in both parts is radial and that the induced charges on the facing flat faces are equal and opposite. Then I assumed that since the gap between the faces was negligible, the potential throughout the sphere is constant, and thus the system is equal to a conducting sphere having a charge Q distributed evenly over it.

Am I correct in assuming thus?

How can the force of interaction be found in that case?

Hellec408 said:
We are given a conducting solid sphere, and it is cut by a plane which has a minimum height r/2 from the centre of the sphere, which has radius r.
I think this is saying that the perpendicular distance from the center of the sphere to the plane is r/2. Is that correct?

Hellec408 said:
A charge Q is given to the smaller part of the conductor, and it is required to find the induced charge on the curved and flat surfaces of the other hemisphere, and also the force of interaction between the two parts of the sphere.
The plane divides the sphere into two unequal-sized pieces. Neither piece is a hemisphere. (I probably sound too nit-picky, but I want to make sure that I'm interpreting the problem statement correctly.)

Hellec408 said:
My initial approach was to apply Gauss law, as we know that electric field is always perpendicular to the conductor surface, and hence we can claim that field in both parts is radial
Yes, the field is radial just outside the curved surface of either piece. But there is also an electric field in the gap between the two flat surfaces. This field is not radial.

Hellec408 said:
and that the induced charges on the facing flat faces are equal and opposite.
Yes.

Hellec408 said:
Then I assumed that since the gap between the faces was negligible, the potential throughout the sphere is constant,
Yes. In crossing the electric field in the gap, the change in potential will be negligible since the distance between the surfaces is infinitesimal.

Hellec408 said:
and thus the system is equal to a conducting sphere having a charge Q distributed evenly over it.
Yes, this is correct for the charge on the outer surfaces of the two pieces. But, you cannot neglect the charge on the flat surfaces when finding the net force on one of the pieces. The oppositely charged flat surfaces attract each other.

Have you found the amount of charge on the curved and flat surfaces of the larger piece? These can be expressed in terms of Q and r.

Hellec408
Thanks, got it!

Last edited by a moderator:
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