As shown in the figure,a square insulating slab 5.0 mm thick measuring 2.0 m × 2.0 m has a charge of 8.0 × 10^-11 C distributed uniformly throughout its volume.Use Gauss's law to determine the electric field at point P,which is located within the slab beneath its center,1.0 mm from one of the faces.
(ε₀ = 8.85 × 10^-12 C²/N • m²)

Question

As shown in the figure,a square insulating slab 5.0 mm thick measuring 2.0 m × 2.0 m has a charge of 8.0 × 10<sup>-11</sup> C distributed uniformly throughout its volume.Use Gauss's law to determine the electric field at point P,which is located within the slab beneath its center,1.0 mm from one of the faces. (ε<sub>0</sub> = 8.85 × 10<sup>-12</sup> C<sup>2</sup>/N • m<sup>2</sup>) A) 0.68 N/C B) 14 N/C C) 23 N/C D) 34 N/C E) 57 N/C

Quizplus · Accepted Answer

Using Gauss's law, we can find the electric field at point P by creating a Gaussian surface that surrounds P and intersects the slab.

We can choose a cube as the Gaussian surface with a side length of 1 mm (since point P is 1 mm from one of the faces of the slab). The cube has a volume of (0.001 m)^3 = 1 × 10^-9 m^3.

Since the charge is distributed uniformly throughout the volume of the slab, the charge density is given by:
ρ = Q/V = (8.0 × 10^-11 C) / (2.0 m × 2.0 m × 5.0 × 10^-3 m) = 4.0 × 10^-7 C/m^3

Using Gauss's law, we have:
∫E·dA = Qinside / ε0
E ∫dA = ρ V / ε0
E (6A) = (4.0 × 10^-7 C/m^3) (1 × 10^-9 m^3) / (8.85 × 10^-12 C^2/N·m^2)
E = 0.68 N/C

Therefore, the electric field at point P is 0.68 N/C.

As Shown in the Figure,a Square Insulating Slab 5