Force on charges separated by grounded spherical condcutor

AI Thread Summary
The discussion revolves around calculating the force on a charge positioned outside a grounded conductive spherical shell. The initial assumption was that the force could be calculated simply using F=(q^2)/(4d^2), but this neglects the influence of the conductive sphere. The presence of the sphere causes charge redistribution, leading to additional electrostatic effects that must be considered. The correct approach involves accounting for the induced charges on the sphere's surface, affecting the force calculation. There is uncertainty regarding the final expression's dimensional consistency, but it aligns with the expected charge distributions.
Leah_Sh
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Homework Statement


Grounded conductive spherical shell is given. Radius R. Charge is held in d and (-d).
d>R. as in the pic.
What is the force on the right charge?

Homework Equations


F=q*q/r^2


The Attempt at a Solution


I thought it's just F=(q^2)/(4d^2).
Why it's wrong?
the right answer is marked.
Tnx.
 

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Your suggestion would be correct if there were no conductive sphere there. However, the sphere has some effects. The presence of the positive charge to the right will electrostatically pull some negative charge to the rightmost surface of the sphere, and similarly on the leftmost surface of the sphere (except opposite polarity). That is the source of the extra terms at a distance of d-R and d+R respectively.

That said, I am unsure about the final expression, it doesn't quite seem dimensionally consistent but it is the only answer which seems to have charges in the right spots with the right polarities.
 

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