Question

a metal plate of thickness half the separation between the capacitor plates of capacitence C is inserted.what will be the new capacitence?

Answer 1

im not sure if i am reading this right but if you put a conductor (metal) in a capacitor where the dielectric (glass) is supposed to be , i wold have thought the charge would not build up at all

Answer 2

but the ans given to me is 2C ....i dont know how this is the ans

Answer 3

@UnkleRhaukus

Answer 4

the capacitance Increases?

Answer 5

yes from the answer it seems so

Answer 6

A conductor actually NEGATES the electric field inside it (E inside a conductor is zero).
So what you are doing is basically reducing the separation between the plates by half.


As capacitance of a parallel plate capacitor = epsilon*A/d (d = separation)
reducing separation by half leads to doubling of capacitance

Answer 7

@FoolAroundMath thnx

Answer 8

@FoolAroundMath this is still not making sense in my head,

the capacitance is the energy stored in the electric field between the plates, right?
the conductor cannot store electric field energy, right?

take the extreme case,
if the conductor a wide enough to be touching both plates , woulden't all the charge flow through without any build up what-so ever hence the Capacitor would have no capacitance.
?

Answer 9

I generally use the definition of capacitance to be the charge stored per unit potential difference.
Suppose the charge stored is constant. Electric field produced by the plates is constant.

So, the total energy stored inside the region = work done by the electric field inside the region between the plates = E*l

Now in the region where there is a conductor, E =0, thus the energy stored in that region = 0. Thus, we have less energy and consequently less potential difference.
Thus, more amount of charge gets stored with less P.D. as before resulting in an increase in the capacitance.
I don't know exactly about touching the plates. You are supposed to be storing them. If you touch, then it wont exactly be static electricity and I don't know how to get around that. On the other hand, if the thickness of the conductor infinitesimally differs from the separation, then indeed if you fix the potential difference. Then V = E*l, l-->0
So E which is proportional to charge density is proportional to V/l --> infinity