What is the Electric Potential at a Point on the Surface of a Charged Sphere?

AI Thread Summary
The discussion centers on calculating the electric potential at a point on the surface of a charged sphere influenced by an external point charge. It is noted that the induced charges on the sphere result in a net potential of zero due to equal positive and negative charges. However, the potential due to the external charge is initially calculated as 6 volts, while the correct answer is 3 volts, suggesting a misunderstanding of the charge distribution. The concept of image charges is introduced as a method to simplify the problem by replacing the sphere with an equivalent point charge, but there is debate over its necessity. Ultimately, the potential at the surface of the sphere is affected by the non-uniform distribution of induced charges.
utkarshakash
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Homework Statement


A point charge q is placed at a distance 2r from the centre O of a conducting charged sphere of radius r. Due to the induced charges on the sphere, find the electric potential at point P on the surface of sphere(it is at a distance 3r from q) (if Kq/r = 18 volt)

Homework Equations



The Attempt at a Solution


Due to the induced charges the potential will be 0 as there are as many -ve charge on the surface of sphere as that of +ve charge. So the net potential is due to q which comes out to be 6 but correct answer is 3.
 
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This is best handled by image methods. Given the location of q outside your sphere, where and what size& polarity is the image charge inside the sphere? Then you dump the sphere and work with just the two point charges.
 
rude man said:
This is best handled by image methods. Given the location of q outside your sphere, where and what size& polarity is the image charge inside the sphere? Then you dump the sphere and work with just the two point charges.

How can there be a net charge on the sphere when it is initially neutral? It must be zero. Half of the sphere will contain -ve charge and the other half, +ve charge.
Also what do you mean by image methods?
 
As I said, once the image charge is in place the sphere is gone. So there is no charged sphere or any other kind.

The function of the image charge is to geneate a field that is indistinguishable from the charge-redistributed sphere at the point of observation. The so-synthesized field is invalid for regions inside the sphere since the field inside a pure conductor has to be zero.
 
rude man said:
As I said, once the image charge is in place the sphere is gone. So there is no charged sphere or any other kind.

The function of the image charge is to geneate a field that is indistinguishable from the charge-redistributed sphere at the point of observation. The so-synthesized field is invalid for regions inside the sphere since the field inside a pure conductor has to be zero.

I don't think that this simple problem requires me to use such kind of methods.
 
utkarshakash said:

The Attempt at a Solution


Due to the induced charges the potential will be 0 as there are as many -ve charge on the surface of sphere as that of +ve charge. So the net potential is due to q which comes out to be 6 but correct answer is 3.

The number of charges are same but they aren't uniformly distributed due to polarization. They will produce a potential.
 
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