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No, when you connect the sphere to earth, the effective conductor now includes the whole of earth.

The induced charges in a neutral sphere (at (5.5,0,0) with radius 9.5) in a non-uniform external electric field coming from a point charge at (-15,0,0) will have the following potential on the z=0 plane:

The sphere has 0V because the net induced charge on the sphere is 0V plus the initial net charge of the sphere (0V because it is a neutral sphere) so the 0V contour coincides with the centre of the sphere.

When you now superimpose that with the potential due to the external point charge:

The sphere is equipotential but not quite equipotential in the diagram but this is only due to the error of the simulation as there are only 1000 surface charges to improve computation speed and it is also difficult to get them uniformly spaced (Thomson problem).

It can be seen though that the potential from the induced charges at any given point on the sphere exactly cancels the potential difference between a point at the centre of the sphere in the external field and the and the given point, meaning that the entire sphere becomes the potential of the point at the centre of the sphere in the external electric field, this appears to be around 3V in this case.

The sphere has a potential of 3V from the centre to infinity in the presence of the point charge but as soon as the external field from the point charge is taken away, it returns to 0V. The only way to keep |3V| on the sphere is to ground the sphere, here's why.

When you ground the sphere, it makes the conductor the size of earth plus the size of the sphere. Let's just model this as one huge sphere. We can therefore assume that the potential at the centre of the sphere from the external field is effectively zero (as opposed to 3V of before) because it is so far away. We also know that the charges induced on the neutral earth conductor exactly cancel out the potential so that the sphere is equipotential with the centre of the sphere, this means that there needs to be -3V from the induced charges on the whole conductor to cancel out the field on the original sphere part of the conductor, so the original sphere part ends up charged with -3V so that sphere section now becomes 0V within the field but will reveal itself to be -3V when the 3V due to the field is removed when you remove the point charge.

No, when you connect the sphere to earth, the effective conductor now includes the whole of earth.

The induced charges in a neutral sphere (at (5.5,0,0) with radius 9.5) in a non-uniform external electric field coming from a point charge at (-15,0,0) will have the following potential on the z=0 plane:

The sphere has 0V because the net induced charge on the sphere is 0V plus the initial net charge of the sphere (0V because it is a neutral sphere) so the 0V contour coincides with the centre of the sphere.

When you now superimpose that with the potential due to the external point charge:

The sphere is equipotential but not quite equipotential in the diagram but this is only due to the error of the simulation as there are only 1000 surface charges to improve computation speed and it is also difficult to get them uniformly spaced (Thomson problem).

It can be seen though that the potential from the induced charges at any given point on the sphere exactly cancels the potential difference between a point at the centre of the sphere in the external field and the and the given point, meaning that the entire sphere becomes the potential of the point at the centre of the sphere in the external electric field, this appears to be around 3V in this case.

The sphere has a potential of 3V from the centre to infinity in the presence of the point charge but as soon as the external field from the point charge is taken away, it returns to 0V. The only way to keep |3V| on the sphere is to ground the sphere, here's why.

When you ground the sphere, it makes the conductor the size of earth plus the size of the conductor and the wire connecting them. Let's just model this as one huge sphere. We can therefore assume that the potential at the centre of this huge conductor due to the point charge is effectively zero (as opposed to 3V of before) because it is so far away. We also know that the charges induced on the huge neutral conductor (neutral sphere + neutral earth) exactly cancel out the potential so that the huge conductor is equipotential with the centre of the huge conductor, this means that the original sphere becomes 0V (so 3V less than before it was grounded), therefore the sphere itself has a net voltage of -3V but is 0V in the field. Disconnecting ground causes the sphere to remain at -3V (still 0V in the presence of the external charge and then removing the external charge reveals the -3V.

No, when you connect the sphere to earth, the effective conductor now includes the whole of earth.

The induced charges in a neutral sphere (at (-5.5,0,0) with radius 9.5) in a non-uniform external electric field coming from a point charge at (-15,15,0) will have the following potential on the z=0 plane:

The sphere has 0V because the net induced charge on the sphere is 0V plus the initial net charge of the sphere (0V because it is a neutral sphere) so the 0V contour coincides with the centre of the sphere.

When you now superimpose that with the external field:

The sphere is equipotential but not quite equipotential in the diagram but this is only due to the error of the simulation as there are only 1000 surface charges to improve computation speed and it is also difficult to get them uniformly spaced (Thomson problem).

It can be seen though that the potential from the induced charges at any given point on the sphere exactly cancels the potential difference between a point at the centre of the sphere in the external field and the and the given point, meaning that the entire sphere becomes the potential of the point at the centre of the sphere in the external electric field, this appears to be around 3V in this case.

The sphere has a potential of 3V from the centre to infinity in the presence of the point charge but as soon as the external field from the point charge is taken away, it returns to 0V. The only way to keep |3V| on the sphere is to ground the sphere, here's why.

When you ground the sphere, it makes the conductor the size of earth plus the size of the conductor. Let's just model this as one huge sphere. We can therefore assume that the potential at the centre of the sphere from the external field is effectively zero (as opposed to 3V of before) because it is so far away. We also know that the charges induced on the neutral earth conductor exactly cancel out the potential so that the sphere is equipotential with the centre of the sphere, this means that there needs to be -3V from the induced charges on the whole conductor to cancel out the field on the original sphere part of the conductor, so the original sphere part ends up charged with -3V so that sphere section now becomes 0V within the field but will reveal itself to be -3V when the 3V due to the field is removed.

No, when you connect the sphere to earth, the effective conductor now includes the whole of earth.

The induced charges in a neutral sphere (at (5.5,0,0) with radius 9.5) in a non-uniform external electric field coming from a point charge at (-15,0,0) will have the following potential on the z=0 plane:

The sphere has 0V because the net induced charge on the sphere is 0V plus the initial net charge of the sphere (0V because it is a neutral sphere) so the 0V contour coincides with the centre of the sphere.

When you now superimpose that with the potential due to the external point charge:

The sphere is equipotential but not quite equipotential in the diagram but this is only due to the error of the simulation as there are only 1000 surface charges to improve computation speed and it is also difficult to get them uniformly spaced (Thomson problem).

It can be seen though that the potential from the induced charges at any given point on the sphere exactly cancels the potential difference between a point at the centre of the sphere in the external field and the and the given point, meaning that the entire sphere becomes the potential of the point at the centre of the sphere in the external electric field, this appears to be around 3V in this case.

The sphere has a potential of 3V from the centre to infinity in the presence of the point charge but as soon as the external field from the point charge is taken away, it returns to 0V. The only way to keep |3V| on the sphere is to ground the sphere, here's why.

When you ground the sphere, it makes the conductor the size of earth plus the size of the sphere. Let's just model this as one huge sphere. We can therefore assume that the potential at the centre of the sphere from the external field is effectively zero (as opposed to 3V of before) because it is so far away. We also know that the charges induced on the neutral earth conductor exactly cancel out the potential so that the sphere is equipotential with the centre of the sphere, this means that there needs to be -3V from the induced charges on the whole conductor to cancel out the field on the original sphere part of the conductor, so the original sphere part ends up charged with -3V so that sphere section now becomes 0V within the field but will reveal itself to be -3V when the 3V due to the field is removed when you remove the point charge.

No, when you connect the sphere to earth, the effective conductor now includes the whole of earth.

The induced charges in a neutral sphere in a non-uniform external electric field will have the following potential on the z=0 plane:

The sphere has 0V because the net induced charge on the sphere is 0V plus the initial net charge of the sphere (0V because it is a neutral sphere) so the 0V contour coincides with the centre of the sphere.

When you now superimpose that with the external field:

The sphere is equipotential but not quite equipotential in the diagram but this is only due to the error of the simulation as there are only 1000 surface charges and it is difficult to get them uniformly spaced (Thomson problem).

It can be seen though that the potential from the induced charges at any given point on the sphere exactly cancels the potential difference between a point at the centre of the sphere in the external field and the and the given point, meaning that the entire sphere becomes the potential of the point at the centre of the sphere in the external electric field, this appears to be around 3V in this case.

The sphere has a potential of 3V from the centre to infinity in the presence of the point charge but as soon as the external field from the point charge is taken away, it returns to 0V. The only way to keep the 3V on the sphere is to ground the sphere, here's why.

When you ground the sphere, it makes the conductor the size of earth plus the size of the conductor. Let's just model this as one huge sphere. We can therefore assume that the potential at the centre of the sphere from the external field is effectively zero (as opposed to 3V of before) because it is so far away. We also know that the charges induced on the neutral earth conductor exactly cancel out the potential so that the sphere is equipotential with the centre of the sphere, this means that there needs to be -3V from the induced charges on the whole conductor to cancel out the field on the original sphere part of the conductor, so the original sphere part ends up charged with -3V so that sphere section now becomes 0V within the field but will reveal itself to be -3V when the 3V due to the field is removed.

No, when you connect the sphere to earth, the effective conductor now includes the whole of earth.

The induced charges in a neutral sphere (at (-5.5,0,0) with radius 9.5) in a non-uniform external electric field coming from a point charge at (-15,15,0) will have the following potential on the z=0 plane:

The sphere has 0V because the net induced charge on the sphere is 0V plus the initial net charge of the sphere (0V because it is a neutral sphere) so the 0V contour coincides with the centre of the sphere.

When you now superimpose that with the external field:

The sphere is equipotential but not quite equipotential in the diagram but this is only due to the error of the simulation as there are only 1000 surface charges to improve computation speed and it is also difficult to get them uniformly spaced (Thomson problem).

It can be seen though that the potential from the induced charges at any given point on the sphere exactly cancels the potential difference between a point at the centre of the sphere in the external field and the and the given point, meaning that the entire sphere becomes the potential of the point at the centre of the sphere in the external electric field, this appears to be around 3V in this case.

The sphere has a potential of 3V from the centre to infinity in the presence of the point charge but as soon as the external field from the point charge is taken away, it returns to 0V. The only way to keep |3V| on the sphere is to ground the sphere, here's why.

When you ground the sphere, it makes the conductor the size of earth plus the size of the conductor. Let's just model this as one huge sphere. We can therefore assume that the potential at the centre of the sphere from the external field is effectively zero (as opposed to 3V of before) because it is so far away. We also know that the charges induced on the neutral earth conductor exactly cancel out the potential so that the sphere is equipotential with the centre of the sphere, this means that there needs to be -3V from the induced charges on the whole conductor to cancel out the field on the original sphere part of the conductor, so the original sphere part ends up charged with -3V so that sphere section now becomes 0V within the field but will reveal itself to be -3V when the 3V due to the field is removed.