Electric field is zero in the center of a spherical conductor

[anuttarasammyak]

I don't understand how charge in a spherical conductor is only on the surface unless induced by a field.

If there are charges under surface, there should be electric field inside the conductor. Free electrons move so that these charges inside disappear.

 

[Orodruin]

Again, there are two lines of argumentation going on here and I become more and more convinced that the OP is mixing them up.

- The field in a hollow central region of a spherically symmetric charge distribution is zero. This is based on the distribution of charges alone. It has nothing to do with the sphere being a conductor.

- The field inside an ideal conductor (or just a conductor without externally applied potential that has reached a stationary state) is zero. This is based on the resistivity of an ideal conductor being zero and has nothing to do with the conductor being a sphere.

Until the OP clarifies that he understands the difference between those two and specifies which is intended, I do not believe we will get any further.

 

[vanhees71]

Again, there are two lines of argumentation going on here and I become more and more convinced that the OP is mixing them up.

- The field in a hollow central region of a spherically symmetric charge distribution is zero. This is based on the distribution of charges alone. It has nothing to do with the sphere being a conductor.

As shown in my posting above, it's 0 everywhere inside the sphere, if there's a spherically symmetric charge distribution, that is vanishing inside this sphere. It's indeed irrelevant, whether the sphere is a conductor or not.

 

[Orodruin]

As shown in my posting above, it's 0 everywhere insight the sphere, if there's a spherically symmetric charge distribution, that is vanishing inside this sphere.

Well, this is what I said already in my first post but people did not like the use of ”inside” … I have some doubts whether or not the integration argument will satisfy the OP who seems more intent on qualitative reasoning than actually mathing it out … the confusion regarding what the actual question is does not help either …

 

[Alex Schaller]

Electric field is 0 in the center of a spherical conductor. At a point P (black dot), I do not understand how the electric field cancels and becomes 0. Electric field is in blue.
View attachment 300530

Feynman puts it simple, in his "Lectures on Physics", page 5-5:

If the spherical shell is conductive, charge would spread evenly. At any point within the sphere, take for instance "P", the field is due to the contribution of the charges on the sphere. If you think of two opposite cones at "P", each one contributes with the charge that is on the base of that cone (intersecting the sphere). This charge is equal to charge density times area of intersection. The area increases with the square of the distance to "P". The field due to each charge decreases with the square of the distance as well. So they both cancel resulting in a same field contribution for any cone, so if you take cones by pairs (the field of one cone is equal and opposite of the field of the other cone) and integrate over the whole sphere, it yields zero field at any point within the sphere.

 

[annamal]

Again, there are two lines of argumentation going on here and I become more and more convinced that the OP is mixing them up.

- The field in a hollow central region of a spherically symmetric charge distribution is zero. This is based on the distribution of charges alone. It has nothing to do with the sphere being a conductor.

- The field inside an ideal conductor (or just a conductor without externally applied potential that has reached a stationary state) is zero. This is based on the resistivity of an ideal conductor being zero and has nothing to do with the conductor being a sphere.

Until the OP clarifies that he understands the difference between those two and specifies which is intended, I do not believe we will get any further.

I understand that a solid spherical conductor with charge on its surface has 0 electric field inside. I am asking why this conductor also have 0 electric field if charge is uniformly inside the sphere.

 

[annamal]

Feynman puts it simple, in his "Lectures on Physics", page 5-5:
View attachment 300615
If the spherical shell is conductive, charge would spread evenly. At any point within the sphere, take for instance "P", the field is due to the contribution of the charges on the sphere. If you think of two opposite cones at "P", each one contributes with the charge that is on the base of that cone (intersecting the sphere). This charge is equal to charge density times area of intersection. The area increases with the square of the distance to "P". The field due to each charge decreases with the square of the distance as well. So they both cancel resulting in a same field contribution for any cone, so if you take cones by pairs (the field of one cone is equal and opposite of the field of the other cone) and integrate over the whole sphere, it yields zero field at any point within the sphere.

If there were charge inside both cones, the electric field of the right cone would cover some of the electric field in the left cone leaving extra electric field in the left cone.

 

[Orodruin]

Feynman puts it simple, in his "Lectures on Physics", page 5-5:
View attachment 300615
If the spherical shell is conductive, charge would spread evenly. At any point within the sphere, take for instance "P", the field is due to the contribution of the charges on the sphere. If you think of two opposite cones at "P", each one contributes with the charge that is on the base of that cone (intersecting the sphere). This charge is equal to charge density times area of intersection. The area increases with the square of the distance to "P". The field due to each charge decreases with the square of the distance as well. So they both cancel resulting in a same field contribution for any cone, so if you take cones by pairs (the field of one cone is equal and opposite of the field of the other cone) and integrate over the whole sphere, it yields zero field at any point within the sphere.

This is in essence the same argument I made in #23 with the additional pointing out that the area grows as the distance squared (which it has to in order to cancel). However, there is also an additional point that must be made to ensure the validity of the argument and it is that both cones intersect the sphere’s surface at the same angle because the area is not proportional only to the distance squared, but also to the reciprocal of the cosine of the intersection angle. This is a however a bit of elementary geometry that any straight line through a circle (or sphere) will cross the circle at the same angle at both crossings.

If there were charge inside both cones, the electric field of the right cone would cover some of the electric field in the left cone leaving extra electric field in the left cone.

I am sorry to say, but based on your posts here, you seem to have fundamental misunderstandings regarding what the electric field is and how it is applied in this situation. The electric fields from the charges inside both cones are equal in magnitude and opposite in direction just by the argument provided. This means they exactly cancel.

 

[Orodruin]

I understand that a solid spherical conductor with charge on its surface has 0 electric field inside. I am asking why this conductor also have 0 electric field if charge is uniformly inside the sphere.

It doesn’t. But if it is a conductor, there will not be charge uniformly distributed inside the sphere.

 

[Alex Schaller]

If there were charge inside both cones, the electric field of the right cone would cover some of the electric field in the left cone leaving extra electric field in the left cone.

If the cones are within a conductor, they would contain no (unbalanced) charge; all the charges would try to separate as much as they can (because they are repelling each other) until they reach the surface of the conductor (here they cannot proceed farther as the air surrounding the conductor is an isolator - for low voltages). Only an isolating material could have charge inside.

 

[vanhees71]

I understand that a solid spherical conductor with charge on its surface has 0 electric field inside. I am asking why this conductor also have 0 electric field if charge is uniformly inside the sphere.

As I stressed already several times, there IS a non-vanishing field, if there is charge inside the sphere!

 

[annamal]

As I stressed already several times, there IS a non-vanishing field, if there is charge inside the sphere!

What do you mean by non vanishing?

 

[Orodruin]

What do you mean by non vanishing?

Non-zero …

 

[annamal]

Non-zero …

Ok, I see. Thank you for your patience with me. I get it now.

 

[sophiecentaur]

In a conductor that is a solid sphere without any external electric field, there should be charge uniformly throughout the sphere..

I understand that a solid spherical conductor with charge on its surface has 0 electric field inside. I am asking why this conductor also have 0 electric field if charge is uniformly inside the sphere.

It doesn’t. But if it is a conductor, there will not be charge uniformly distributed inside the sphere.

Take a conductor. Charges will always flow to the lowest potential in a conductor. The lowest potential arrangement in a solid sphere is with all the charges on the surface (mutual repulsion with lowest potential arrangement when they are spaced around the surface). The only sort of sphere with uniform charge in it would be a perfect insulator with the charges locked in evenly spaced locations. You actually don't get a lot of them.