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Could anyone explain why the intensity of the electric field between plates of a charged capacitor is constant? Moreover, the varying the distance between plates doesn't change the electric field intensity - that's weird, because the electric field is defined as the force acting on a unit charge, and the force according to Coulomb law certainly does depend on the distance between the charges.

So it's reasonable to expect that placing two plates further apart would result in a lower electric field intensity (or, lower force experienced by a unit charge placed between the plates), but that's not the case (for some reason).

See example 4.2 if you need technical details.

Let's say the value of charge, not potential, is fixed on both plates. Does the electric field insensity change with the distance between the plates? The answer has to be 'no', because doubling the distance between plates of a capacitor doubles the voltage across them (and $V=Ed$). And if the electric field intensity remains constant (it's just force acting on unit charge), then the force acting on the test charge will be the same no matter how far apart the plates are.

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  • $\begingroup$ They assume capacitors as infinitely wide, or its distance is negligibly small compared to its area. $\endgroup$ – philip_0008 Jul 18 '16 at 17:49
  • $\begingroup$ It depends on whether you are keeping the potential constant, or not. If you keep the charge constant, then the field will stay the same and the potential will increase, but if you keep the potential constant, then the charge will decrease and the field will be smaller. $\endgroup$ – CuriousOne Jul 18 '16 at 19:24
  • $\begingroup$ @CuriousOne yeah, and if I keep the charge constant, and increase the distance, the electric field will stay constant - why? If I place a test charge half a distance between the plates, and write down the electric force acting on it, and then move each plates 100m away from the charge, how can it be that the force acting on the test charge will remain unchanged? That's the case if the field stays the same, right? Isn't it a bit counterintuitive? Where did the Coulomb law go? $\endgroup$ – user4205580 Jul 18 '16 at 19:52
  • $\begingroup$ The electric field stays constant because the charge density on the surface stays constant. $\endgroup$ – CuriousOne Jul 19 '16 at 0:22
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For an INFINITE parallel plate capacitor, the electric field has the same value everywhere between the 2 plates. An intuitive reason for that is: suppose you have a small test charge +q at a distance $x$ away from the +ve plate and a distance $d-x$ away from the -ve plate. The +ve plate will repel the charge and the -ve plate will attract it. Now if the charge is at a distance $y$ from the +ve plate $(y>x)$ then the the repulsive force due to the +ve plate will be weaker but the attractive force due to the -ve plate will be stronger such that the net force on the test charge will have the same value as when the charge was a distance $x$ away from the +ve plate. This is because the electric field lines due to an infinite plate are parallel to the normal vector to the plate.

I don't quite understand your second question. If you have a fixed potential difference between the two plates, then the electric field between the plates is given by $E=\frac{V}{d}$, where d is the distance between the 2 plates. From the expression, you can see that the electric field is NOT independent of the separation of the 2 plates.

Now suppose you keep the charge on both plates the same and separate them, such the the distance, $d$ increases. In that case, the electric field will remain unchanged. This might seem counter intuitive at first sight, this might seem like a violation of the conservation of energy. But, the extra energy required to keep the E field the same comes from the work that you are putting in to separate the attracting plates. The capacitance, $C$=$\frac{\epsilon A}{d}$, where A is the area of the plates, d is the separation of the plates and $\epsilon$ is the permittivity of the material between the plates. If $\epsilon$ and $A$ are constants, then, $C$ is inversely proportional to $d$.

The relation between the potential difference between the 2 plates, $V$ and the charge on one of the plates, $Q$ is given by $Q=CV=\frac{\epsilon A V}{d}=\epsilon A E$.

$\implies E=\frac{Q}{\epsilon A}$

Therefore, E is independent of $d$ if the charge on each plate is unchanged.

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  • $\begingroup$ Now let's say the value of charge, not potential, is fixed on both plates. Does the electric field insensity change with the distance between the plates? The answer has to be 'no', because doubling the distance between plates of a capacitor doubles the voltage across them (and $V = Ed$). $\endgroup$ – user4205580 Jul 18 '16 at 18:04
  • $\begingroup$ I have edited my answer to include the case of constant charges on each plate. $\endgroup$ – PhysLab Jul 18 '16 at 21:09
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    $\begingroup$ What about Coulomb's law? If I place a test charge half a distance between the plates, and write down the electric force acting on it, and then move each plates 100m away from the charge, how can it be that the force acting on the test charge will remain unchanged? What about Coulomb's law? We have charged bodies (plates) and test charge. The force acting on the charge should depend on the distance, but it doesn't. $\endgroup$ – user4205580 Jul 18 '16 at 21:13
  • $\begingroup$ The reason why you are probably confused is that you are thinking about finite sized plates, in which case you would be right. But for a charged infinite plate, the electric field lines are straight and parallel to each other. They extend to infinity. $\endgroup$ – PhysLab Jul 18 '16 at 21:27
  • $\begingroup$ The only components of the force on the test charge that do not cancel out are in the direction parallel to the normal to the plate. You will see that the other components cancel out if you think about it. When the test charge is close to the plate, even though the force on the test charge due to each point charge in the plate is stronger that if the test charge was far apart, the NET force will be the same since the components parallel to the normal of the plate will be smaller. Try drawing a diagram of the forces and check which components of the forces cancel, it will become clear. $\endgroup$ – PhysLab Jul 18 '16 at 21:33
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A capacitor is supposed to have infinite dimensions.And Electric field strength of a charged plane sheet of infinite dimensions is constant over infinity i.e. distance does not matter.

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Using Gauss theorem we measure electric flux i.e. $$ \phi=E.dA $$ and if you use Gauss theorem for this particular geometry of plane parallel plates then you will get the usual result where the distance from the plate of the point where the field is to be measured will not appear. Further, E is proportional to $ \frac{1}{r^2}$ and the area $dA$ is proportional to $r^2$ and while taking the dot product the dependency over the distance is cancelled out. Actually, if you consider a infinite dimensional pill box as the surface for using the Gauss theorem for plate geometry then also you will get only the term the charge enclosed by the surface. Though the surface becomes infinite charge at the plate surface will not change the flux will also be independent of the distance.

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protected by Qmechanic Jul 21 '16 at 11:18

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