Electrostatic shielding in a cavity

[@bhishek]

Suppose a constant electric field exists in space as shown in the file and a open lid cube type conductor is brought inside the field. What would be the electric field in the question marked area. Would it be zero or equal to original electric field.

 

[TESL@]

It would be zero.
File clean: https://www.virustotal.com/en/file/...e321fae9d85b910a9c5f16c3/analysis/1418672218/

 

[@bhishek]

It would be zero.
File clean: https://www.virustotal.com/en/file/...e321fae9d85b910a9c5f16c3/analysis/1418672218/

Why??

 

[TESL@]

Electric fields cancel out inside of a conductor.

This is a good visual: http://en.wikipedia.org/wiki/Faraday_cage#mediaviewer/File:Faraday_cage.gif

 

[@bhishek]

Why??

Electric fields cancel out inside of a conductor.

This is a good visual: http://en.wikipedia.org/wiki/Faraday_cage#mediaviewer/File:Faraday_cage.gif

But the Faraday's cage is a closed one. Will the same principal be applicable to an open lid conductor also??
Thank You for the replies.

 

[TESL@]

Lid being open would slightly matter if it faced the electric field vectors, that the charge would accumulate on the tips and cause a curved electric field. The net electric field would still approach to zero as you got closer to the bottom. But in this case, it is definitely zero.

 

[@bhishek]

If that is the case than two conductors of any shape having an electrical path between them will have nil electric field between them when placed inside an electric field. Can we stretch the logic that far??

 

[TESL@]

Not that far. Think of a hollow semi-sphere and a small sphere on its center, connected to it. When you apply an electric field in center to semi-sphere direction, there will be perpendicular fields between the end of the semi-sphere and the center, non-negligible.

 

[@bhishek]

Not that far. Think of a hollow semi-sphere and a small sphere on its center, connected to it. When you apply an electric field in center to semi-sphere direction, there will be perpendicular fields between the end of the semi-sphere and the center, non-negligible.

Thank you,for explaining.
Is this type of shielding used anywhere?

 

[TESL@]

The regular Faraday cage is used but I don't know about the last device, I just made it up.

 

[@bhishek]

The regular Faraday cage is used but I don't know about the last device, I just made it up.

Thanks once again.