In the above picture, I near positively rod to a neutral sphere. Then I ground the right side (negative side). Why electrons enter to sphere from ground?
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2$\begingroup$ Duplicate of : physics.stackexchange.com/q/492264/207455 $\endgroup$– user207455Commented Jul 18, 2019 at 16:55
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$\begingroup$ in this situation i near positive rod to sphere. then negative side connect to ground.when connect negative side to ground, why electron go up. electrons repel each other. how this occur? $\endgroup$– Reza RezaeiCommented Jul 18, 2019 at 17:00
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$\begingroup$ Did you mean that you bring the positively charged rod close to the neutral spere without actual contact? $\endgroup$– user207455Commented Jul 18, 2019 at 17:03
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$\begingroup$ yes. then grounding right side( negative side). $\endgroup$– Reza RezaeiCommented Jul 18, 2019 at 17:04
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$\begingroup$ Then you should clarify the question as it is difficult to follow. $\endgroup$– user207455Commented Jul 18, 2019 at 17:07
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2 Answers
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Initially these conducting spheres are neutral bodies but when a positively charged conductor is brought near to them, the charges inside the sphere are redistributed on the surface of the sphere in such a way that the charges have minimum repulsions as you have shown.
But due to the positively charged rod, the spheres aquaire some potential. For general idea of what is happening during earthing let us consider a point charge $+Q$, then initial potential of sphere due to charge $+Q$ is $$V_{i}=kQ/r$$ where r is distance from positive charge $+Q$ to the sphere.
When two conductors of different potentials are connected by a conducting wire, charge transfer takes place between the conductors until they attain same potential.
We know that Earth is also a conductor with huge number of charges and almost negligible potential due to it's large radius (for reference we consider it to be zero). So when the sphere is earthed, charge transfer takes place between Earth and the sphere until the potential of the sphere becomes zero.
In the case you have shown, negative charges flow from Earth to the conductor so that this negative charge cancels the potential on the sphere due to positive charge of the rod and bring the sphere to zero potential.
Let $q_{e}$ be the negative charge that flows from Earth to sphere and $V_{f}$ be the final potential of the sphere which equals to zero. R be radius of the sphere.
We express it as:
$$V_{f}=kQ/r + kq_{e}/R = 0$$
Here $kQ/r$ is potential on the Sphere due to positively charge and $kq_{e}/R$ is potential on the sphere due to negative charges that flow from earth to conductor. They cancel out each other which brings the sphere to zero potential.
So to bring the sphere to zero potential (reference to Earth), the electrons enter from ground to sphere.
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When positively charged rod is brought near sphere all electrons come on side near the rod. other side then becomes positively charged. when the sphere is connected to ground electrons flow from ground to sphere to neutralize the positive charge.
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$\begingroup$ right side of sphere grounded not left side. $\endgroup$ Commented Jul 20, 2019 at 12:06
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$\begingroup$ it does not matter which side you ground. see physics.stackexchange.com/questions/102311/… $\endgroup$– user193922Commented Jul 21, 2019 at 8:23