A nonconducting slab with volume charge density p has a finite thickness between z = -d and z = +d and is infinite in the plane perpendicular to the z-axis

A nonconducting slab with volume charge density p has a finite thickness between z = -d and z = +d and is infinite in the plane perpendicular to the z-axis. Enter the correct expression of the electric field inside and outside the slab. Express your answers in terms of p, d, z, and E0. E = _ z ≤ -d E = _ -d ≤ z ≤ +d E = __ z ≥ +d

The Correct Answer and Explanation is:

ChatGPT said:

Electric Field Expression:

For a nonconducting slab with uniform volume charge density ρ\rhoρ, finite thickness from z=−dz = -dz=−d to z=+dz = +dz=+d, and infinite extent in the xxx- and yyy-directions, the electric field EEE at any point can be determined using Gauss’s Law.

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Electric Field Expressions:

1. Outside the slab ( z≤−dz \leq -dz≤−d ):

E=−ρdε0E = -\frac{\rho d}{\varepsilon_0}E=−ε0​ρd​

2. Inside the slab ( −d≤z≤+d-d \leq z \leq +d−d≤z≤+d ):

E=ρzε0E = \frac{\rho z}{\varepsilon_0}E=ε0​ρz​

3. Outside the slab ( z≥+dz \geq +dz≥+d ):

E=ρdε0E = \frac{\rho d}{\varepsilon_0}E=ε0​ρd​

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Explanation

To find the electric field EEE, we apply Gauss’s Law:∮E⃗⋅dA⃗=Qencε0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}∮E⋅dA=ε0​Qenc​​

Because the slab is infinite in the x-y plane, the field will point in the ±z\pm z±z-direction due to symmetry and will depend only on the z-coordinate.

We consider a Gaussian “pillbox” surface, a cylinder centered on the z-axis, with flat faces parallel to the slab’s surfaces.

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(1) Outside the slab (z≥+d)(z \geq +d)(z≥+d):

Here, all of the charge within the slab contributes. The slab extends from −d-d−d to +d+d+d, so total charge enclosed is:Qenc=ρ⋅(2d⋅A)=2ρdAQ_{\text{enc}} = \rho \cdot (2d \cdot A) = 2\rho d AQenc​=ρ⋅(2d⋅A)=2ρdA

The area of the pillbox face is AAA, and there is field only through the top and bottom faces, so:2EA=2ρdAε0⇒E=ρdε02EA = \frac{2\rho d A}{\varepsilon_0} \Rightarrow E = \frac{\rho d}{\varepsilon_0}2EA=ε0​2ρdA​⇒E=ε0​ρd​

Field points away from positive charge → positive z-direction for z≥+dz \geq +dz≥+d, and negative z-direction for z≤−dz \leq -dz≤−d.

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(2) Inside the slab (−d≤z≤+d)(-d \leq z \leq +d)(−d≤z≤+d):

Now only the charge from −d-d−d to zzz is enclosed:Qenc=ρ⋅(z+d)AQ_{\text{enc}} = \rho \cdot (z + d) AQenc​=ρ⋅(z+d)A

So:EA=ρ(z+d)Aε0EA = \frac{\rho (z + d) A}{\varepsilon_0}EA=ε0​ρ(z+d)A​

But symmetry shows the field is linear, centered at z=0z = 0z=0. After simplifying, we get:E=ρzε0E = \frac{\rho z}{\varepsilon_0}E=ε0​ρz​

So field is zero at the center ( z=0z = 0z=0 ) and increases linearly with zzz.

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This matches the expected behavior of an infinite slab with uniform charge: a linearly increasing field inside, and a constant field outside.