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Suppose, you have to capacitor plates as shown:

enter image description here

And, according to the solution key to a problem which I was doing, in the steady state behaviour of the circuit, the charges would redistribute such that each plate gets charge of equal magnitude. Like so:

enter image description here

Now, this was the method taught to us for solving problems where extra charge is introduced into capacitors but I'm not sure why the charges redistributed such that each plate has average of both plates? Why does at end state the charge redistribute so that each plate has the same amount of charge

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3 Answers 3

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If you provide a path, the electrons may move from/to a plate.

If there is no path, the electrons have to remain on a plate.

You have not drawn any paths.

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I think you may have created an anomaly in the way you have drawn your picture. Capacitor plates that are parallel and of the same size will have equal and opposite charge. But for there to be unequal charge on two capacitor plates there needs to be a difference in the plate areas and, this creates "fringing" to a third party (usually ground) like this: -

enter image description here

The overlap area between left plate and upper right plate will have +500 and -500 values for charge. The lower right plate (representing the rest of the universe) will have +200 and -200 charge values. You could also redraw it like this: -

enter image description here

But, by definition of a capacitor, it is a device that HAS equal and opposite charges on its plates meaning that the +200 charge surplus on the +700 plate has to produce leakage flux to other stuff. This means that if the "other stuff" is a much greater distance away than the two larger plate's gap, then the net average voltage on the two larger plates can be quite large with respect to the "other stuff".

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This is actually a simple result of Gauss Law. Consider a cylindrical closed surface whose cross section is shown below (in dotted lines):

enter image description here

Due to the symmetry of the plates the electric field is pointing in direction perpendicular to the surface of the plates as shown in the figure. So how much flux crosses the cylindrical surface? The answer is zero because the electric field is pointing along the surface in air/dielectric and electric field is zero inside the conductor (of-course assuming electrostatic conditions). From Gauss's Law,

$$\Phi_E = \frac{q_{enc}}{\epsilon_o}$$ Since \$\Phi_E\$ is zero the net charge enclosed inside the cylinder is zero. This is only possible if the charges on the two plates are equal and opposite. The final charge configuration is thus, as shown below: enter image description here

Note that inner surface of the plates have equal and opposite charges and outer plates carry the residual charges.

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