 #1
yucheng
 213
 57
 Homework Statement:

See https://physics.stackexchange.com/questions/275950/uniquenesstheoreminuniformelectricfield
The sphere is an equipotential we can set it to zero. Then by symmtery the entire $xy$ plane is at potential zero. Then $V$ does not go to zero but rather far from the sphere the field is $E_0 \hat{z}$ we thus have $$v \to E_0z + C.$$
 Relevant Equations:
 N/A
The answer given states that:
The entire xy plane is obviously at the same potential since all the fields are strictly perpendicular to it (draw a diagram if youre confused). Since we choose the sphere to be at potential zero, the point on the sphere which cuts the xy plane is also at zero potential, and hence the entire plane is at zero by the definition chosen
My concerns are:
1. Which xy plane? ##z=0## only, or ##R \leq z \leq R##?
2. Won't the argument thus apply to the whole 'slab' with thickness ##2R## (along zaxis, diameter of sphere), and therefore along the whole slab, it would be of zero potential?
Essentially, why will Griffiths claim that 'The sphere is an equipotential we can set it to zero. Then by symmtery the entire $xy$ plane is at potential zero. ' ?
Thanks in advance!
The entire xy plane is obviously at the same potential since all the fields are strictly perpendicular to it (draw a diagram if youre confused). Since we choose the sphere to be at potential zero, the point on the sphere which cuts the xy plane is also at zero potential, and hence the entire plane is at zero by the definition chosen
My concerns are:
1. Which xy plane? ##z=0## only, or ##R \leq z \leq R##?
2. Won't the argument thus apply to the whole 'slab' with thickness ##2R## (along zaxis, diameter of sphere), and therefore along the whole slab, it would be of zero potential?
Essentially, why will Griffiths claim that 'The sphere is an equipotential we can set it to zero. Then by symmtery the entire $xy$ plane is at potential zero. ' ?
Thanks in advance!