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S is a solid neutral conducting sphere. ...

S is a solid neutral conducting sphere. A point charge q of 1 × 10–6 C is placed at point A. C is the centre of sphere and AB is a tangent. BC = 3m and AB = 4m.
(1) The electric potential of the conductor is 1.8 kV
(2) The electric potential of the conductor is 2.25 kV
(3) The electric potential at B due to induced charges on the sphere is – 0.45 kV
(4) The electric potential at B due to induced charges on the sphere is 0.45 kV

A

The electric potential at B due to induced charge on the sphere is 1.2 kV

B

The electric potential at B due to induced charge on the sphere is –1.2 kV

C

The electric potential at B due to induced charge on the sphere is –0.45 kV

D

The electric potential at B due to induced charge on the sphere is 0.45 kV

Text Solution

Verified by Experts

The correct Answer is:
C
The potential of a conducting sphere is constant at all points of its body so, we have
`V_(B)=V_(C)`, where `V_(C)=(Kq)/(R),R=5m` ( `:.` potential due to induced charges at C is zero)
At point B potential is due to the charge at A and due to the induced charges on the sphere so we have
`V_(B)=V_(i)+(Kq)/(r),r=4mimpliesV_(i)=(K_(q))/(R)-(K_(q))/(r)=-0.45kV`
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