Shield Principle Clarified: Inside-Outside Electric Field Independency

[babblingsia]

Consider an uncharged spherical conductor .It has a cavity carved out of it and there's a charge q in the cavity.Now you get an electric field outside the sphere.All fine.But why does Feynman say that "...no static distribution of charges inside a closed conductor can produce any fields outside.Shielding works both ways!In electrostatics-but not in varying fields- the fields on the two sides of a closed conducting shell are completely independent." Could someone please help me clear the inconsistencies?

 

[siddharth]

Consider an uncharged spherical conductor .It has a cavity carved out of it and there's a charge q in the cavity.Now you get an electric field outside the sphere.All fine.

Yeah, that's right.

But why does Feynman say that "...no static distribution of charges inside a closed conductor can produce any fields outside.Shielding works both ways!In electrostatics-but not in varying fields- the fields on the two sides of a closed conducting shell are completely independent." Could someone please help me clear the inconsistencies?

That' doesn't seem right. Are you sure you that's exactly how it was given in his book, and you didn't make an error while posting?

If so, he must talking about grounded conductors, or conductors which has a net opposing charge.

 

[Doc Al]

Feynman error!

Yes, Feynman really did write that just as babblingsia stated. It's an error that has been corrected in the latest printing of the "Definitive" edition. It now reads:

...the fields on the two sides of a closed grounded conducting shell are completely independent.​

PS: When I dragged out my tattered copy of the lectures to verify the OP's quote, I noticed the notation "huh?" that I had penciled in next to that statement when I first read it decades ago.

 

[Meir Achuz]

"the fields on the two sides of a closed conducting shell are completely independent."
I think he means that moving a charge wilthin a cavity (thus changing the field in the cavity) does not affect the field outside the conductor.
Sometimes Feynman's colorful language can confuse.

 

[pardesi]

i feel what he means is the the charge distribution inside the cavity has absolutely no effetc what so ever on the outside field

 

[rahuldandekar]

Yep. See the example in Griffiths Introduction to ED, section on conductors. Very enlightening.