Frequently in physics, one makes simplifying approximations. A common one in ele

Question

Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion of infinite charged sheets. This approximation is useful when a problem deals with points whose distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you will look at the electric field from two finite sheets and compare it to the results for infinite sheets to get a better idea of when this approximation is valid. Consider two thin disks, of negligible thickness, of radius R oriented perpendicular to the x axis such that the x axis runs through the center of each disk. (Figure 1) The disk centered at x = 0 has positive charge density eta, and the disk centered at x = a has negative charge density -eta, where the charge density is charge per unit area. What is the magnitude E of the electric field at the point on the x axis with x coordinate a/2? Express your answer in terms of eta, R, a., and the permittivity of free space Element_0. E = 4 pi kn(1 - a/Squareroot a^2 + R^2)

Explanation / Answer

Ans:-

First find the field due to the +disc.
The field dE ,at arbitrary, but constant x, due to point charge dq in the disc at radius r is;
dE = kdq/(r^2 + x^2)

From symmetry, only Ex will be non zero so you only need to find that component;
dEx = dECos@
= (dE)x/SR[r^2+x^2]
= kxdq/{r^2+x^2)^3/2

Write charge dq as;
dq = ndA = nrdrd(pi)

Integrate immediately over phi from 0 to 2pi;
Ex = 2(pi)knxINTEGRAL[rdr/(r^2+x^2)^3/2)]
Integrate this from 0 to R
Ex = 2(pi)kn[1 - x/SR(R^2+x^2)]
evaluate at x=a/2
Ex = 2(pi)kn[1 - a/SR(a^2+4R^2)]

The field due to the negative disc will have exactly the same magnitude at a/2 and its direction will also be up, so the two fields just add to give.

Ex = 4(pi)kn[1 - a/SR(a^2+4R^2)]
k = 1/4(pi)0

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