Electric flux density a direct measuer of E?

[dE_logics]

Temp message

This is what a book says -

“The number of lines per unit area through a surface perpendicular to the lines is
proportional to the magnitude of the electric field in that region. Thus, the field
lines are close together where the electric field is strong and far apart where the
field is weak.”
+
“These properties are illustrated in Figure 23.20. The density of lines through
surface A is greater than the density of lines through surface B. Therefore, the magni-
tude of the electric field is larger on surface A than on surface B. Furthermore, the fact
that the lines at different locations point in different directions indicates that the field
is nonuniform.”

What I concluded from the above -

That means the total intensity of E.F passing though an area is independent of the number of lines of force, depends more its density and is directly proportional to it.

So even if we are comparing 2 areas having areas A and 2A and suppose x lines of forces passes through them so the total intensity of the field at A should be greater than 2A, if the areas are charged partially, then the force on A will be more than 2A despite the fact that the number of lines of forces passing through both of them are the same the E.F on A will be higher cause the density of the lines are A is higher.
Talking about insane analogies, Suppose we have 2 area A and 500000*A, and if the density of the lines in A is more than 500000*A, the force on A will be more than on 500000*A. Suppose the the flux density on A is 1000 lines of forces/m² and on 500000*A is 999 lines of forces/m²...then also by what the book says, A will experience more force/field.

Considering the above, suppose we have 2 areas -

http://img223.imageshack.us/img223/5449/2areas.jpg

Then even if the small area is one trillionth of the larger area, it will experience more force simply cause the flux density is more on it.

Edit:Consider the charge on both the areas as the same.

Well...sounds insane to me.

And what do you mean when one line of force passes through an area?...I mean the field at this state should have a minimum possible value, but still needs to decrease with an inverse square relation to distance...what will happen then?

 

[dx]

The fact that the electric field on A is greater doesn't mean the force on A is greater. Assuming E points in the same direction everywhere on the plate, the force on A will be equal to σAE where σ is the surface charge density.

 

[dE_logics]

The fact that the electric field on A is greater doesn't mean the force on A is greater. Assuming E points in the same direction everywhere on the plate, the force on A will be equal to σAE where σ is the surface charge density.

Ok...I've changed the question a bit, I've included the fact that the total charge on each of the areas are the same.

 

[dx]

Let's say the total charge on each plate is Q, distributed uniformly.

If the area of plate one is A, then the charge density on it will be σ₁ = Q/A. If the area of plate two is 2A, then the charge density on it will be σ₂ = Q/(2A).

If E is the same on both plates, then the force on both plates will be the same because EA^Q/_A = E⋅2A⋅^Q/_2A = EQ. If E is different on the two plates, say E₁ and E₂, then the forces will be E₁Q and E₂Q respectively.

 

[dE_logics]

Ok, I thought about it -

Suppose we have 2 areas...A and xA...where x is a positive integer greater than 1.
On xA and flux density of y falls, while on A a flux density of ay falls...where a is a positive integer greater than 1.

The net charge on both these areas is the same...q.

What I mean to say here, is that suppose we have 2 areas one larger, one smaller, the flux density on A is greater than xA and the charges on both the areas is the same.
So according to the fact that flux density is a direct measure of strength of the field, what will the force on A and xA?

The charge (and so force) on both these areas will be by virtue of certain charged particles uniformly distributed on both these area, since the amount of charged particles on both these areas is the same, the final thing that will matter is the flux density, since that will define force falling on each particle...and the net force will be the summation of all these forces, which's the same for both the areas.

We can replace the area with these charges and get the same effect, force is directly proportional to the flux density.

 

[dE_logics]

Thanks for telling me about that charge on the areas.

 

[dx]

What I mean to say here, is that suppose we have 2 areas one larger, one smaller, the flux density on A is greater than xA and the charges on both the areas is the same.
So according to the fact that flux density is a direct measure of strength of the field, what will the force on A and xA?

I already answered this question. If E on plate one is E₁ and E on plate two is E₂, then the forces are QE₁ and QE₂.

 

[dE_logics]

Yes, I didn't see that, I was making that...but you posted it before.

Thanks!

 

[dE_logics]

Is this true for magnetic fields also?