As shown in the figure, a square insulating slab 5.0 mm thick measuring 2.0 m x 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. (0 = 8.85 × 10-12 C2/N • m2)

Principles of Physics: A Calculus-Based Text
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Chapter19: Electric Forces And Electric Fields
Section: Chapter Questions
Problem 53P: Consider a thin, spherical shell of radius 14.0 cm with a total charge of 32.0 C distributed...
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As shown in the figure, a square insulating slab 5.0 mm thick measuring 2.0 m x 2.0 m has a
charge of 8.0 x 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. (e0 = 8.85 × 10-12 C2/N • m²)
1.00 mm
P.
5.00
1.
mm
2.00 m
O 34 N/C
14 N/C
23 N/C
O 0.68 N/C
O 57 N/C
Transcribed Image Text:As shown in the figure, a square insulating slab 5.0 mm thick measuring 2.0 m x 2.0 m has a charge of 8.0 x 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. (e0 = 8.85 × 10-12 C2/N • m²) 1.00 mm P. 5.00 1. mm 2.00 m O 34 N/C 14 N/C 23 N/C O 0.68 N/C O 57 N/C
Expert Solution
Step 1

Let us the total flux pass through both faces 2A = 2 dx dy of gaussian box 2 dx dy dz.

 

The electric flux is parallel to the z-axis only,

 

According to gauss's law

 

Flux = E.2A

Flux = Qenclosedε0          =ρ. volε0          =E 2dx.dy              =ρ.dx.dy.dzε0          = ρ.dzε0

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