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23 Nov 2019

A parallel plate capacitor with an air dielectric is attached toa voltage source and charged.
The voltage source is removed, and then the plates are broughtcloser together to half their
previous distance. What happens to the electric field between theplates when they are brought
together?
A. It is diminished by a factor of 4.
B. It halves.
C. It doubles.
D. It stays the same.
E. It quadruples.

explain please, it says it stays the same but i got the equation E= vBL , so if L halves then doesn't E also half. What equation doyou use for parallel plate capacitors?

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Sixta Kovacek
Sixta KovacekLv2
7 Apr 2019

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Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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This has been bothering me for a long time and I cant find anadequate explanation..

Lets say you have an infinite conducting plate with finitethickness D and it is given charge Q

the charge will be equally split into two and concentrated only onthe two surfaces.

Each surface will get Q/2

and the magnitude of electric field outside the plate isE=Q/(2A*epsilon naught) (E for a conducting plate is surface chargedensity/epsilon naught as opposed to surface chargedensity/(2epsilon naught)

Then, if you have two nonconducting plates separated by distance Dand each plate gets Q/2 which is uniformly distributed ..
then outside the space between the two plates, E = 2*(Q/4A*epsilonnaught)

But why cant we use superposition principle in the case of oneconducting plate with Q and say you get E=Q/(A epsilon naught)i.e.Q/(2A epsilon naught) from each surface...

Is it because we are looking at the mid section of the conductor assome kind of Faraday cage? (since the plate is essentially infinityand blocks the electric field from the other side?)

Also, lets look at two parallel plate capacitor..

One plate gets +Q/2 and the other gets -Q/2

I know the formula says E= charge density/epsilon naught.. so E hasto be Q/(2A epsilon naught) However, it seems rather inconsistentin the case of the parallel plate capacitor...

We know all the charge is going to be concentrated on the innersurface
So shouldnt you get E = Q/(2A epsilon naught) solely from one plateand since you have TWO plates!!!, shoudlnt you multiply Q/(2Aepsilon naught) by 2?

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