Question

A planar slab of thickness of 5.00 cm has a uniform volume charge density of 7.20×10^-2 C/ m³. Find the magnitude of the electric field at all points in space both inside and outside the slab, in terms of x, the distance measured from the central plane of the slab. What is the field for x = 1.25 cm?

What is the field for x = 10.00 cm?

Answer 1

thickness of the slab is t =5 cm =0.05 m, charge density ρ = 7.20×10^-2 C/ m^3

at x1 distance from the the central plane of the slab the amount of charge Q =2*x1*A*ρ

A is area of the Gaussian surface,

according to the Gauss's law, flux φ=Q/ε0

but flux φ=2*E*A (on both sides)

therefore 2*E*A = Q/ε0

E =(2*x1*A*ρ)/(2*A*ε0) = x1ρ/ε0

now electric feild at any point inside the slab at a distance x from the central plane of the slab is

E =xρ/ε0

here ε0=8.85*10^-12 C^2/N m^2

therfore electric feild at point x =1.25 cm =0.0125 m

E =(0.0125)*(7.20e-2)/(8.85e-12)

E = 1.01 e 8 V/m

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at x2 distance from the the central plane of the slab the amount of charge Q =d*A*ρ is charge enclosed by the slab

here A is area of the Gaussian surface

according to the Gauss's law

flux φ=Q/ε0

but flux φ=2*E*A (on both sides)

therefore 2*E*A = Q/ε0

E =(d*A*ρ)/(2*A*ε0)

electric feild at any point outside the slab at a distance x from the central plane of the slab is

E =d*ρ/ε0

therefore E =d*ρ/2ε0

=(0.05)*( 7.2e -2)/(2*8.85e-12)

E = 2.03 e 98V/m