As shown in the figure, a square insulating slab 5.0 mm thick measuring 2.0 m As
ID: 2261305 • Letter: A
Question
As shown in the figure, a square insulating slab 5.0 mm thick measuring 2.0 m As shown in the figure, a square insulating slab 5.0 mm thick measuring 2.0 m times 2.0 m has a charge of 8.0 times 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 epsilon 0 = 8.85 times 10-12 C2/N middot m2) As shown in the figure, a square insulating slab 5.0 mm thick measuring 2.0 m times 2.0 m has a charge of 8.0 times 10-11 C distributed uniformly throughout its volume. Use Gauss's law to determine the electric field at point exttip{P}{P} P which is located within the slab beneath its center, 1.0 mm from one of the faces. ( arepsilon_0 epsilon 0 = 8.85 times 10-12 V2/N middot m2)Explanation / Answer
Place origin of coordinates at the center of the slab
and let the total flux pass though both faces (2A=2*d_x*d_y) of
Gaussian box (2*d_x*d_y*d_z) such that the electric flux is
parallel to the z-axis only.
By Gauss's law and symmetry:
Flux = E.2A = Q_enc/?o
Flux = E2A = p*Vol / ?o
Flux = E2(d_x*d_y) = p*(d_x*d_y*2*d_z)/?o
=> E = p*d_z/?o
Data:
Let permittivity of slab be approximately:
?o=8.85*10^-12 F/m
Distance from origin to point in question:
d_z=0.0015 m
Uniform charge density of slab:
p = Q/vol = (8 * 10^-11)C / (0.005 * 2 * 2)m^3
=> E = p*d_z/?o = 0.68 N/c
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