## Sunday, June 13, 2010

### Irodov Problem 3.44

In this problem first we need to find the electric field due to an infinite thin plane sheet with a hole. For this, we shall first use the superposition principle. The idea being that the electric field due to a sheet with cavity is same as the summation of fields due to a disc with negative charge density and a solid infinite plane sheet with positive charge density as shown in figure 1.

First let us determine the electric field due to a charged disc. Consider an infinitesimally thin ring section on the disc (as shown in Figure 2) of thickness dr and radius r.The surface area of this ring will be . The charge contained in this infinitesimally thin ring will thus be . We have already derived the expression for the electric field due to a charged ring at a distance x along the axis from Problem 3.9 Equation 4. Using this expression we have the electric field dE due to this infinitesimally thin ring as

(1) thus, gives the electric field due to a positively charged disc at a distance x along the axis of the disc.

Now let us determine the elxtric field due to an infinitely large thin charged sheet. For this we can use Gauss law. Consider a Gaussian surface that is in the shape of a cylinder with its axis perpendicular to the sheet and having and area of cross-section A. The total charge contained within this Gaussian cylinder is . The electric field E will be perpendicular to the Gaussian cylinder at both its ends. Hence we have,

The electric field due to the infinite sheet with a circular hole can now be determined as,

Now using (3) we can determine the electric field due to two such sheets, one negatively charged and the other positively charged separated by a distance of l from each other (as shown in Figure 4) as,

When l is very small we can use Taylor's series to approximate (4) as,

When (5) becomes,

Now the electric field potential can be determined as,

When we have,

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I just got the result that electric field is sigma/2epsilonnought (i.e. constant) for an infinite plane at any point in space, so I decided to duck, and it seem that you've goth same result. Any insight on why this happens? Shouldnt the field decrease as a function of perpendicular difference?

ReplyDeleteThe field is constant because if you try to consider the angle subtended by the top of the plate at any point near the surface, it is always 90°. Any change in separation causes a negligible change in this angle, as the prospect is much larger than the separation - think of a right angled triangle with base 100000000000000 m and the altitude changing from 10m to 20m. Thus, the field remains constant as long as the plate is infinite - that is, as long as the distance from the plate is small compared to its size. If the plate is strictly infinite, this obviously holds for all finite distances.

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