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When a neutral conductor is positively charged , the excess charges move to the boundary. Initially, inside the conductor there exist a field due to these excess charges which as per my reading results in the rearrangement of mobile electrons to cancel the initial field. My doubt is how exactly the electrons are redistributed without leading to another electric field inside the conductor.

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    – uhoh
    Aug 4, 2017 at 14:31

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The "NET" Electric field inside a "conductor" is always zero as you have said.This is basically due to the presence of a lot of free electrons in the conductor.

Let us consider a conducting plate kept in a uniform electric field perpendicular to the plane of the conductor as shown.

enter image description here

The electric field exerts a force F=-eE on the the free electrons of the conductor pulling the electrons to the left side ( to the direction opposite to the electric field) of the conductor creating negative charge accumulation on that side. This in-turn creates a positive charge accumulation on the other side of the conductor. This accumulation starts to create an electric field opposite in direction to the existing electric field ( that is the new electric field opposes the already existing one). This accumulation increases and so do the opposing electric field. This continues until both the electric field become equal in magnitude. So the new opposing electric field completely cancels out the already existing on (only inside the conductor). So the net electric field in inside the conductor becomes 0. All this happens basically in nano seconds. So, the conductor isn't affected by any new electric field coming across it as it is cancelled out instantly. enter image description here

So, Getting to your question, the excess charges creates an electric field inside the conductor.Now,the mobile electrons rearranges as you have said to create another electric field which will totally cancel out the existing electric field inside the conductor. Its the "net electric field" that will be zero inside a conductor.

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In this case, the conductor is positively charged, meaning that the freely moving electrons rush towards the center of the conductor to fill in the positively charged gaps. Looking at this process in slow motion (because it's all happening very fast), we'll see that all the gaps will gradually be filled, Leaving a positive charge density on under the boundary of the conductor. The electric field due to this positive charge density on the boundary will be zero everywhere inside the conductor (Gauss'law), below the boundary.

So, on the boundary, a positive shell will form. If we consider a conductor with finite spatial dimensions, then the electric field produced outside the conductor will approach that of a positive point charge (with a charge depending on the number of holes inside the conductor) if we move to infinity (inversely proportional to the squared distance to the point). If the conductor has two finite and one infinite spatial dimension, the field produced outside the conductor will approach that of a positively charged line (inversely proportional to the distance to the line). And when the conductor has one finite and two infinite spatial dimensions the electric field outside the conductor will approach a field produced by a positively charged infinite plane (independent of the distance to the plate).

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I'm new to this subject, and I'm here because I was having the same question myself, which I will consider to be about a spherical charged conductor, and without reference to an exogenous electric field.

It is obvious that if all excess charge is distributed uniformly around the surface, the field at the center will be zero. But I was concerned that at interior points very close to the surface the field couldn't be uniformly zero, because if it was zero at some point, p, at some tiny distance from some point on the surface, s, it couldn't also be zero at the point, h, half-way between p and s, because s would exert a (4x) stronger force at h than it did on p, while other points on the surface would have only marginally different influences on p and h, leading to an imbalance of forces. However, it occurs to me that my concern is unfounded in a 3-D world.

This is because in the limit locally, the surface of a sphere becomes a plane, and it is a theorem that the electric field exerted by a plane does NOT depend on the distance between the test charge and the plane. So in fact, s (or a small fixed 2-D neighborhood around it) exerts the same force on both p and h, and so the net force at p and h can be the same.

Notice, though that this argument seems to show that the field will not be zero everywhere for a 2-D or 1-D sphere, where the force does in fact increase as a test charge approaches the "surface", ( a line in the case of a 2-D "sphere" or a point for a 1-D "sphere".)

Extra Credit Thought Question: What happens in the case of a 4-D sphere?

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