Charge Distribution on a perfectly conducting hollow shell As shown in the figure, I have a hollow shell which contains a charge at its centre and another charge is placed outside the shell (some distance apart).

I know that the situation this figure depicts is NOT in electrostatic equilibrium as, of course the field inside the conducting material is not zero, the charges will flow in such a way that the resulting field due to induced charges will exactly CANCEL out the field due to other charges.

What is don't understand is that inside the sphere (the region which is not filled with conducting material) there will be no contribution of the field due to the charge placed outside the shell (I was told this by my instructor) so the force on charge placed at centre is zero.

I can't deduce a reason for this fact. I want to understand why this happens. Any help will be appreciated.

• Please use tags properly. Why did you add the "conformal-field-theory" tag? You should click on the tag and know what it's about before using it. – harshit54 Jan 14 at 16:17
• Oops I must have added it by mistake! I'll remove it – Shivansh J Jan 14 at 16:22

You are familiar with the electric potential of a field? Then here is how it goes:

• since the final configuration is in equilibrium, the field has to be perpendicular to the surface of the sphere for every surface point. More specific the tangental component has to zero at every surface point (otherwise there would be a force and therefore acceleration on the surface charges, F=Eq)

• this implies that the potential is constant on the surface (just connect any two points on the surface and calculate the line integral of the E field)

• this in turn implies the potential is constant inside the sphere. Think of it this way, if you have any straight path from one side to the other side inside of the sphere, the total integral needs to be zero. If the E field wouldnt be zero at each point then the integral would be positive at some point, turning negative and at this point gauss law will tell us there has to be a charge (this is an intuitiv argument, mathematically we know that the laplace operator on a surface of potential zero has only zero as a solution)

• if the potential is constant 'everywhere' the electric field (the gradient) is zero

• YESSS!! Thank you very much! I'm delighted to finally find a satisfactory answer to my query. – Shivansh J Jan 14 at 17:37
• @lalala Are the arguments which I gave in my answer, correct? – harshit54 Jan 15 at 12:32
• @harshit54 I find it difficult to follow. Why do you think one has to add up energy in pairs? – lalala Jan 15 at 13:31
• @lalala Sorry for the late reply. For finding the energy of a system we either add all the pairs OR we bring them one by one to the final positions and add up the energies. Both lead to the same answer. Here is video about it. – harshit54 Jan 15 at 16:06
• @harshit54 I probably understand why 3. must vanish. But this does not explain why the sum of 1. 2. and 3. is at a minimum, or does it? – lalala Feb 20 at 13:37

This type of a set up is called a Faraday's Cage and this phenomenon is called electrostatic shielding.

The charge placed outside should have had some electric field on the inside. However, the shell will distribute its charge in such a way that the electric field inside the non-conducting region cancels out and therefore the charge on the inside does not feel any force from the outside.

Fun Fact:

This is also the reason why you should stay inside a car if the weather is stormy and lightning is present in the sky.

Why do they do it?

Because charges inside a conductor are free to move and moving in this way will reduce the total energy of the system.

How is energy reduced?

I can't give you an exact math expression for showing this but can give you an intuitive idea.

You must know that for finding the energy of a system, we sum up the energies between all possible pairs. Here 3 pairs are present

1. Shell & External Charge
2. Shell & Internal Charge
3. Internal & External Charge

If the charge did not distribute, you will be adding the sum of energies of all these pairs.

However if the charge distributes in this way then the 3rd pair will cancel out(because if there is no field then there is no energy).

• I understand that the field will cancel out inside the conducting region. But why does the field (only due to charge placed outside the shell, not the NET field) will cancel out in the region inside the shell which is not filled with conducting material. – Shivansh J Jan 14 at 16:20
• I meant the region inside the shell which is not filled with any material have no electric field. I edited the answer. Does it make sense? – harshit54 Jan 14 at 16:25
• But I want to know the reason behind this. Can you explain why those charges distribute in such a way to cancel the Field due to outside charge? – Shivansh J Jan 14 at 16:27
• @ShivanshJ Edited again. – harshit54 Jan 14 at 16:31
• I don't get how they will reduce the energy of the system by distributing in such a way, can you elaborate further? – Shivansh J Jan 14 at 16:31