Question

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x = d and x = -d. The y- and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density p.a. Explain why the electric field due to the slab is zero at the center of the slab (x=0).

b. Using Gauss's law, find the magnitude of the electric field due to the slab at the points 0<= x <= d. Express your answer in terms of the variables p, x, d, and e_0.

d. Using Gauss's law, find the magnitude of the electric field due to the slab at the points x>= d. Express your answer in terms of the variables p, x, d, and e_0.

e. What is the direction of the electric field due to the slab at the pints x >= d.

Answer 1

a) Because there are no charges inside the slab, the electric field is zero according to Gauss's law.

b)  E = ρ x / 2ε₀  

d)   E = σ / 2ε₀,

e) When the charge is positive, the electric field is exiting the plate.

What is Gauss law?

Gauss' law, often known as Gauss' flux theorem in physics, is a law that relates the distribution of electric charge to the resulting electric field.

a) There are no charges inside the slab, so the electric field is zero according to Gauss' law. Another way to look at it is that the charge on one side of the crockery creates an outgoing electric field in the center, and the charge on the other side of the crockery creates a field of equal magnitude but in the opposite direction, so the resulting field is zero.

b) To compute the electric field, use Gauss' law and a Gaussian surface with the base parallel to the faience as the cylinder, such that the scalar product is reduced to the algebraic product.

Ф = ∫ E. dA = qint /ε₀

A is the slab area.

Consider the concept of charge density.

ρ = qint / V

The volume of the slab is equal to the area multiplied by the thickness.

V = A x

qint = ρ A x

The flow is caused by the electric field created by the two sides of the slab.

Φ = E 2A

2E A = ρ A x /ε₀  

E = ρ x / 2ε₀  

Where x ranges from zero to the thickness of the plate, an x = d

d) If x> d

In this situation, all of the charge is contained within the gaussian surface. The link between volumetric density and surface density is investigated.

σ = Q / A

σ = q d / Ad = Q d / V

ρ = Q / V

σ = ρ d

We can observe that the surface charge density is the product of the density voluntarily by plate thickness.

E = σ / 2ε₀  

e) As the charge is positive, it is leaving the plate in the direction of the electric field.

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