A slab of insulating material has a uniform positive charge density ?, as shown

Question

A slab of insulating material has a uniform positive charge density ?, as shown in the figure below. The slab is infinite in the y and z directions.

Derive expressions for the field for the following regions. (Use the following as necessary: ?0, ?, d, and x as necessary.)

(a) the exterior regions (x > d/2)

(b) the interior region of the slab (0 < x < d/2)

A slab of insulating material has a uniform positive charge density ?, as shown in the figure below. The slab is infinite in the y and z directions. Derive expressions for the field for the following regions. (Use the following as necessary: ?0, ?, d, and x as necessary.) the exterior regions (x > d/2) the interior region of the slab (0

Explanation / Answer

REPLACE z0 BY d/2 in the given solution u will get your answer

Gauss claimed (and I believe him) that the total flux (= field x area) across a closed surface is proportional to the charge inside the volume, E * A = q / e where big E is the electric field (a vector) and little e is the permittivity of the material (since permittivity wasn't given we'll assume the permittivity of free space here, e = 8.85e-12 F/m). The trick is, to choose a useful volume.

(a) Consider a cylinder of radius r with its top at z=+z0 and its bottom at z=-z0. Radially from the axis of the cylinder (i.e., parallel to the xy-plane) it's uniform charge density as far as the eye can see, no matter which direction you look. Since the charge looks the same in every direction, by symmetry the radially-directed E-field at the boundary of the cylinder must be the same in every direction. Now consider a second cylinder just like the first one, but next to it and touching it at a single point. (At a line of points, really, but you know what I mean.) Everything that was true of the first cylinder is true of the second. Even the E-field at the boundary must be the same, because both have the same shape and contain the same charge and are surrounded by exactly the same charge density (if "exactly" is the right word when you're dealing with an infinite extent of anything). Where the two cylinders touch, the E-fields are the same size but pointing in opposite directions. The only way that's not a contradiction is if the radial E-field is zero. Since you can place two such cylinders anywhere in the xy-plane, the E-field parallel to the xy-plane must be zero in every direction. (Wow, that's a lot of words!)

Now suppose that the first cylinder only extends to +/- a in the z-direction, symmetric about the xy-plane. The E-field out of the top of this cylinder must be equal to the E-field out of the bottom of the cylinder, because I can look at the problem upside-down and everything looks the same. The total charge enclosed is
Q = ? (pi r^2) (2a)

and the total flux phi (how do you type a Greek letter?) is
phi = Ez (pi r^2) * 2 (top and bottom, yes, and no radial E out the sides?)

so by Gauss's law
phi = Q / e
Ez (pi r^2) * 2 = ? (pi r^2) (2a) / e
Ez = ? a / e

As a vector,
E = z* ? z / e where z* is a unit vector in the +z direction, and -z0 < z < +z0

(b) Apply the same symmetry argument as in (a) to a cylinder completely above or below the sheet of charge to convince yourself that there is no E-field parallel to the xy-plane outside the charge sheet, either. Now consider a cylinder of radius r that extends upward (or downward - symmetry again) from z=0 to z = b > z0. The cylinder encloses a charge
Q = ? (pi r^2) z0

The E-field at z=0 is zero. If the E-field at z = b is Ez, the flux is
phi = pi r^2 Ez

and Gauss's law says
phi = pi r^2 Ez = Q/e = ? (pi r^2) z0 / e
Ez = ? z0 / e for z > z0

As a vector,
E = z* ? (pi r^2) z0 / e for z > z0
E = - z* ? (pi r^2) z0 / e for z < - z0

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