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Qmechanic
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These days I encoutered the famous grounded sphere near a charge problem, and I saw a pretty straight forward solution(for the image charge induced on the sphere). I am not sure if this solution is OK... so here it is:

Consider a charge "q" at a distance "r" from the center of a grounded ,conducting sphere of radius "R". Find the charge induced on the sphere. (r>R)

Consider a charge $q$ at a distance $r$ from the center of a grounded, conducting sphere of radius $R$. Find the charge induced on the sphere. ($r>R$)

Solution(seen by me)

Because the sphere is grounded it has the potential(inside and on the sphere): $$V=0$$ The charge q' induced on the surfuace of the sphere(and the charge q) ,regardless of how it is distributed , will give in the center of the sphere the total potential: $$V=\frac{q'}{4\pi\epsilon R} + \frac{q}{4\pi\epsilon r} = 0$$ So $$q'=-q \cdot \frac{R}{r} $$

Do you find this ok? If yes , please explain.

(rigurous solution: https://www.youtube.com/watch?v=KoQ3KP2oSMohttp://www.youtube.com/watch?v=KoQ3KP2oSMo)

These days I encoutered the famous grounded sphere near a charge problem, and I saw a pretty straight forward solution(for the image charge induced on the sphere). I am not sure if this solution is OK... so here it is:

Consider a charge "q" at a distance "r" from the center of a grounded ,conducting sphere of radius "R". Find the charge induced on the sphere. (r>R)

Solution(seen by me)

Because the sphere is grounded it has the potential(inside and on the sphere): $$V=0$$ The charge q' induced on the surfuace of the sphere(and the charge q) ,regardless of how it is distributed , will give in the center of the sphere the total potential: $$V=\frac{q'}{4\pi\epsilon R} + \frac{q}{4\pi\epsilon r} = 0$$ So $$q'=-q \cdot \frac{R}{r} $$

Do you find this ok? If yes , please explain.

(rigurous solution: https://www.youtube.com/watch?v=KoQ3KP2oSMo)

These days I encoutered the famous grounded sphere near a charge problem, and I saw a pretty straight forward solution(for the image charge induced on the sphere). I am not sure if this solution is OK... so here it is:

Consider a charge $q$ at a distance $r$ from the center of a grounded, conducting sphere of radius $R$. Find the charge induced on the sphere. ($r>R$)

Solution(seen by me)

Because the sphere is grounded it has the potential(inside and on the sphere): $$V=0$$ The charge q' induced on the surfuace of the sphere(and the charge q) ,regardless of how it is distributed , will give in the center of the sphere the total potential: $$V=\frac{q'}{4\pi\epsilon R} + \frac{q}{4\pi\epsilon r} = 0$$ So $$q'=-q \cdot \frac{R}{r} $$

Do you find this ok? If yes , please explain.

(rigurous solution: http://www.youtube.com/watch?v=KoQ3KP2oSMo)

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CosminA
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Image charge of a grounded sphere

These days I encoutered the famous grounded sphere near a charge problem, and I saw a pretty straight forward solution(for the image charge induced on the sphere). I am not sure if this solution is OK... so here it is:

Consider a charge "q" at a distance "r" from the center of a grounded ,conducting sphere of radius "R". Find the charge induced on the sphere. (r>R)

Solution(seen by me)

Because the sphere is grounded it has the potential(inside and on the sphere): $$V=0$$ The charge q' induced on the surfuace of the sphere(and the charge q) ,regardless of how it is distributed , will give in the center of the sphere the total potential: $$V=\frac{q'}{4\pi\epsilon R} + \frac{q}{4\pi\epsilon r} = 0$$ So $$q'=-q \cdot \frac{R}{r} $$

Do you find this ok? If yes , please explain.

(rigurous solution: https://www.youtube.com/watch?v=KoQ3KP2oSMo)