1.

S is a solid neutral conducting sphere. A point charge q of 1 xx 10^(-6)C is placed at point, C is the center of sphere, and AB is a tangent. BC = 3 m and B = 4m.

Answer»

The ELECTRIC potentialof the CONDUCTOR is 1.8 kV.
The electirc potential of the conductor is 2.25 kV.
The electric potential at B due to induced charges on the sphere is `-0.45kV.`
The electric potential at B due to induced charges on the sphere is 0.45kV.

Solution :a.,c.
Distance `AC` is GIVEN by `AC=5m`
Potential at due to charge `q` at `A` is given by
`V=(kq)/(AC)=(9xx10^(9)xx1xx10^(-6))/(5)=1.8xx10^(3)=1.8kV`
Potential at `B` is given by
`V_(B)=(V_(B))_("due to" q)+(V_(B))_("induced")`
`{(V)(B))_(i)=` Potential at `B` due to induced charge}
THUS `1.8xx10^(3)=(kq)/(AB)+(V_(B))_(i)=2.25xx10^(3)+(V_(B))_(i)`
`(V_(B))_(i)=-0.45kV`
So (a) and (c) are correct


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