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**3.3 Example- Infinite sheet charge with a small circular hole**

Now let’s consider an interesting example that we have an infinitely wide sheet of charge, so it goes to infinity in both of these dimensions. Minus infinity and minus infinity in these directions. Let’s assume that it is positively charged and it has a surface charge density of Sigma Coulombs per meter squared. This distribution has a small circular hole over here with radius r.

We’re interested with the electric field that this distribution generates some z distance away along the axis of the circle at this point p. Therefore e at point p is the question mark.

If we look at this problem and try to solve this problem by applying Coulomb’s law, it’s a very complicated problem. It will generate integrals, specifically they will be very hard to take along this circular region.

Earlier, we did an example by applying Gauss’s law. We calculate an electrical field of an infinite sheet. For an infinite sheet of charge, by applying [pill box] technique, as you remember, we have found that the electric field was equal to, let’s use subscript s over here for the sheet, and that was equal to Sigma over 2 Epsilon zero. This sheet is an insulating sheet of charge.

On the other hand also, we have calculated the electric field of a disc charge with radius r along its axis by applying Coulomb’s law. The disc charge distribution generated an electric field along its axis e sub d. Let’s use subscript d for the disc. That was equal to surface charge density of the disk divided by 2 Epsilon zero times 1 minus z over square root of z squared plus r squared.

Well, if we look at this physical system or distribution, maybe by knowing the results of these types of distributions, or even if we don’t know, that we can easily calculate by applying Gauss’s law for the infinite sheet and Coulomb’s law for the disc charge, that we can find these results and combine these two cases in order to generate the same physical system.

As a matter of fact, we can do that simply first by taking an infinite sheet and we know that the electric field that it generates at a specific point in space, is equal to this quantity over here and it is always constant with the surface charge density of Sigma Coulombs per meter squared. Then we take a disc with radius r and let’s choose this disc with a surface charge density of minus Sigma Coulombs per meter squared. That disc will generate an electric field along its axis z distance away from its center.

Since it is negatively charge in downward direction, since the positive one was in upward direction, if we take this disc with this charge density and superimpose on this distribution, for the sheet of charge, the distance is irrelevant because it always generates Sigma over 2 Epsilon zero of electric field, therefore if we just superimpose this disc on this sheet of charge, then such a system is going to generate a distribution that the area of this disc with a charge density of minus sigma, we’ll neutralize the region of this positively charged sheet of charge generating a region electrically neutral. The remaining parts of this distribution will remain again positively charged with a surface charge density of Sigma.

Having a neutral region over here is going to be equivalent as if we are having a cutout in this sheet of charge. Therefore physically these two systems, this system in our problem and this system will be equivalent to one another. The electric field that this sheet of charge at a location z distance or any distance away from the sheet is positive since it’s positively charged, and it’s pointing in upward direction and the magnitude of that is equal to Sigma over Epsilon zero. This is due to the sheet and this is due to the disc. If we sum these two fields, then we will get the total electric field of this system and that’s what we are after.

Recall the analyzing the problem using this method is superposition, in other words, we superimpose two different systems such that we end up with the charge distribution that we’re dealing with, which is a more complicated case, but we take the advantage of the already known cases or cases that we can easily calculate and solve and superimpose them in order to get the electric field of a more complex distribution.

e total is going to be equal to vector sum of electric field due to the sheet of charge plus due to the disc charge and if we choose our positive direction as upward direction, then the electric field generated by the sheet is Sigma over 2 Epsilon zero and the electric field generated by the disc is going to be minus, since this is going to be in downward direction, Sigma over 2 Epsilon zero times 1 minus z over square root of z squared plus r squared.

Since Sigma over 2 Epsilon zeroes are common terms, we can take it into Sigma over 2 Epsilon zero parentheses and we will have 1 minus z over square root of z squared plus r squared close parentheses. If we take it one more step further, we will have Sigma over 2 Epsilon zero, 1 and minus 1 will cancel, minus minus will make plus, so we’re going to have z over square root of z squared plus r squared as the answer of this distribution.

Again, if we choose the upward direction as positive z direction relative to the x, y, z coordinate system, and the unit vector in that direction is unit vector k along z, we can express this in vector notation by multiplying it with the unit vector in positive z direction.