A solid neutral sphere of radius $R$ is earthened. Now a positively charged body $+Q$ is kept at a distance $r$. Due to induction equal negative and positive charges are accumulated on the surface of the sphere.

Due to the positive charge we brought near the neutral conductor, it gains some potential energy let it be $Vi$.

$$ Vi = kQ/r$$ Now since the earthened conductor and earth(zero potential) are having different potentials, charge flow takes place between them to bring the Sphere to zero potential.

Now how the charges transfer between earth and conductor to make them both at same potential?

I) To balance the excess positive charge +Q which is creating potential on conductor, earth sends some negative charges 'Qe' to conductor.

'Vf' be the final potential of the conductor, i.e zero.

Vf = kQ/r + k(Qe)/R = 0

Q/r + (Qe)/R = 0

Therefore Qe = -QR/r

We got the amount of negative charge that is transferred from earth to the conductor.

But my lecturer is saying "same amount of positive charge will flow from conductor to earth"

Why this happens? Since in a metal lattice positive charges does not move and only negatively charged electrons moves. How can an earthed conductor transfer it's positive charges to earth?

How the actual transfer of charges takes place in a neutral earthen conductor when a positive or negative charge is brought near to it?


If a positively charged object is brought near to (but does not touch) a grounded conducting sphere, electrons from the earth will be attracted to the sphere. They will distribute themselves on the sphere (closer to the positive charge) in a manner which cancels the field from the positive charge inside of the sphere. The sphere will come to zero potential (relative to the earth). Outside the sphere, the potentials will be complex and not described by your formulas which only apply to spherically symmetrical distributions of charge (and assume that the potential is zero at infinity). Without contact, there will be no flow of positive charge to the sphere.


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