Electrical Force between plates(capacitor)

1. haleycomet2

29
we know that electrical field strength,E,can be counted by equation
1. E=V/d,or
2. E=Qq/4$$\pi$$$$\epsilon$$r 2
In the circuit with capacitor, the potential difference ,V is decided by battery.So from the first equation ,the E is not affected by permittivity(dielectric material) but it does affected according to second equation.
What is the correct explanation of the electrical force between plates??

Last edited: Jan 28, 2011
2. Delphi51

3,410
Welcome to PF. Quite an interesting question! Apparently inserting a dielectric between the plates decreases the E field between the plates, so E = V/d does not apply . . . or you can say that an opposing E field due to the separation of charges in the dielectric is subtracted from the V/d one. See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html#c3.

There is something wrong with your second formula - it has the dimensions of a force rather than an electric field.

3. gomunkul51

275
This is an Electric Field around a point charge: E = k(q/r²)

Go back and check where you got this formula: E = V/d and how it is derived.

* Basically the epsilon is in the Electric Potential (V). V for point charge is V = - integral(E dl)

4. haleycomet2

29
I went to check the derivation of the formula E=V/d.Referring to the graph,the work done to move charge A to B can be expressed by:
W=qEd or W=qVAB
Therefore by eliminating q,VAB=Ed,so E=VAB/d

Besides,i also think other derivation,in the circuit consist of capacitor,using the formulas below:
E=$$\sigma$$ /$$\epsilon$$
C=$$\epsilon$$ A/d
Q=CV
$$\sigma$$ =Q/A

E=$$\sigma$$ /$$\epsilon$$
=(Q/A)/$$\epsilon$$
=(CV/A)/$$\epsilon$$
=($$\epsilon$$ AV/Ad)/$$\epsilon$$
=V/d

Derivation above showed that the electric field strength is independent of $$\epsilon$$,dielectric material!?If according to the field strength formula E=kq/r2,it seems that the field strength is independent of capacitor area?!
ps:The charge Q counted from Q=CV is the amount of charge on one side of plate or both side of the capacitor?

Thank you again...=)

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Last edited: Jan 31, 2011
5. Delphi51

3,410
E=kq/r² does not apply directly to parallel plates. It is for point or spherical charges.

Q is the charge on each plate.

Your argument for E = V/d is quite convincing. Wikipedia
http://en.wikipedia.org/wiki/Dielectric
says a dielectric reduces the field and increases the charge, but in another article
http://en.wikipedia.org/wiki/Capacitor
they use E = ρ/ε which suggests to me that increasing charge density and increasing ε would leave E unchanged.

Anyway, how would you measure E inside a material?

6. haleycomet2

29
Oh,thanks ,this is what i think now:
The presence of the dielectric will cause the electric field become weaker,so more charges are transferred to plates until reach the initial electric field strength(same with battery's voltage)again.Therefore,the charge density on the plate is increased,but the electric field(E=V/d)is still the same.

Is it??

can I measure the electric field strength by passing a wire between the plates and measure the force exert on the wire(the flowing electron actually) ,or the effect is too tiny to be observed?

7. Delphi51

3,410
The wire trick won't work. There is no net charge on a wire with current flowing. Anyway, how would you get the wire through the dielectric?

Your latest theory about the E field may well be right. I don't know.