Help me understand an equation for electric field

[the_jackal]

Force on Charge Q

I read this equation in a book:

The force between the plates of a capacitor set up by charges +Q and -Q on opposite plates depends on the electric field E set up by the charges. This field is
E = Q/(εA)

F = (Q/2)*E = Q²/(2εA)
where F is the force
E is the electric field
Q is the charge

How did F = (Q/2)*E ?

 

[masadi]

Force on Charge Q

What book u read?

E is introduced by charge on each plate.
F is interaction E and charge on opposite plate or E that's working on charge.

i think F=Q*E where Q is charge lays on field area another charge

not F = (Q/2)*E

Added after 1 seconds:


What book u read?

E is introduced by charge on each plate.
F is interaction E and charge on opposite plate or E that's working on charge.

i think F=Q*E where Q is charge lays on field area another charge

not F = (Q/2)*E

 

[qyinn]

Re: Force on Charge Q

Actually, the fomula F = (QE)/2 is correct.
I don't remember very well the explanation, but it has to do with the fact that the electric field doesn't have the same value on the 2 sides of one plate. If you think of one of the plates, on one side (towards the interior of the capacitor) the field is E and on the other side (towards the exterior) it is zero.

In the general case, if the fields on the 2 sides of a plate are E1 and E2, the the total force acting on that plate is
F = Q(E1+E2)/2

I'll try to see if I can find a better explanation of this.

 

[dazhen]

Re: Force on Charge Q

First of all, you need to know how the total electric field is calculated.

Let's consider one plate with positive charge +Q and area A. With Gauss law, you can easily get the magnitude of electric field E+=Q/2εA. The factor 2 in the denominator comes from the magnitude of electric field times both the upper area and down area.

Same way can be followed to obtain the electric field of negative charge -Q plate.

When we put two plates together, the electric field gets strong inbetween and cancels outside the double plates. This can be obtained by simple vector superposition. Refer the cartoon attached.

Now we know the total field is Etotal=(E+)+(E_)=Q/εA

When we are asked to compute the force on one plate, we only consider the electric field generated from another plate since the charge can't self-drag itself. For instance, the force on the positive plate is the electric field generated from negative plate multiplying the positive charge, which is E_*Q=(Etotal/2)*Q=(Q/2εA)*Q. That answers your question.

 

[the_jackal]

Re: Force on Charge Q

The book is Microsystem Design, equation 6.7 on page 128.

The factor of 1/2 arises from the fact that E is not the field at the charges. If we imagine that the charge at the surface of the plate occupies a thin layer then the field will vary from zero at the inner boundary of the layer to E in the space outside the plate. The average field acting on the surface charges is E/2 which is where the 1/2 factor comes from.

Hope this helped.

 

[danrop]

Re: Force on Charge Q

I'm not sure I understand (or even agree with) the previous explanation - but here's a way to do it using energy considerations (since we're dealing with a conservative field, there's a useful relationship between potential energy and force)

With a seperation y between the plates, the potential energy of the capacitor is
U = (1/2)*(QV) = (1/2)*(Q²*y)/(2*εA)
where the symbols have the usual meaning and V = Qy/(εA) for a parallel plate capacitor

If we increase the seperation b/w the plates by dy, the new value of potential energy is

U + dU = (1/2)*Q²(y+dy)/(2εA)

so, the increment in P.E. is obviously

dU = (1/2)*Q²dy/(2εA)

and the force F = dU/dy (the force-energy relationship I was talking about)
yielding the experession for F.

 

[the_jackal]

Re: Force on Charge Q

danrop, a course in high school physics will get you the proof of the following:

F = Q²/(2εA)

But this not what i was debating. My question was how does the following equation come about since at high school i was always taught that F = Q*E. But in the equation below you see a mysterious 1/2 factor.

F = (Q/2)*E

How the factor of 1/2 arises is what i tried to explain in my post. Let me know if you need further explaination.

 

[tantoun2004]

Re: Force on Charge Q

The formula u have is correct The explanation is as follows:
*First u can derive it using principle of virtual work and for a harder problem see Feynman Vol II ch 10
*Physically:
On the 2 conductiong plates of the capacitors the charges are in one atomic layer then inside the conductor there is no field so the electric field varies from zero inside the conducting plates to E=σ/ε inside (σ is the surface charge density) so the field AT the charge is not E but rather an averge of E and zero which is E/2
so the force is F=Qσ/2ε
reagrds,

 

[the_jackal]

Re: Force on Charge Q

My solution listed above is actually taken from Feynmen's Lecture which i pointed out to the author whose book contained the equation but provided a very poor explaination to its derivation.

 

[bzaki]

Force on Charge Q

Is there a more detailed explanation for this issue than Feynman's lectures?

Regards,
Bassem

 

[the_jackal]

Re: Force on Charge Q

bzaki, as far as i know there is no detailed explaination to this other than Feynman's lectures. But i believe his work has done a good job in explaining this.

 

[omid_azmoon]

Force on Charge Q

IT'S EASY . HALF OF CHARG IS IN THE LEFT PLATE AND ANOTHER HALF CHARG IS IN RIGHT PLATE . THEN THE FOMULA GO ON CORRECT !