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Asked Oct 14, 2019
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3. An infinite, insulating slab or with uniform charge density p and thickness d is at the
origin. Two infinite sheets of charge density o are each placed a distance d from the origin.
a. Find E(x )
σ
3d
b. Find E (x
42
5d
c. Find E(x
11
4
x= -d
x =0 x= +d
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3. An infinite, insulating slab or with uniform charge density p and thickness d is at the origin. Two infinite sheets of charge density o are each placed a distance d from the origin. a. Find E(x ) σ 3d b. Find E (x 42 5d c. Find E(x 11 4 x= -d x =0 x= +d

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2 Ratings
Step 1

The expression for the Electric field intensity due to infinite sheet of charge is calculated from the integral form of the Gauss’s Law for electrostatics.

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Venc

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Step 2

Here qenc is the charge enclosed by the sheet, E is the electric field vector, dA is the differential area vector, εo is the permittivity of free space.

Given the surface charge density σ of the infinite sheet of charge the, charge enclosed will be,

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Яeтe — ОА вnс

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Step 3

For an infinite sheet of charge the electric field line and the area vector are parallel to each other, and the field lines point in both direction away from...

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Е фал- ОА Е(24) - 9A о3 i (forx<0) E =- E =i(forx>0) 280

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