Respuesta :
Answer:
(a) Surface charge density is the charge per unit area.
[tex]\sigma = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]
(b) [tex]\vec{E} = (+\^z)~8.34\times 10^3~{\rm N/C}[/tex]
(c) [tex]\vec{E} = (-\^z)~8.34\times 10^3~{\rm N/C}[/tex]
Explanation:
(a) Surface charge density is the charge per unit area. The area of the square plate can be calculated by its side length.
[tex]A = l^2 = (0.4)^2 = 0.16 ~{\rm m^2}[/tex]
Half of the total charge is distributed on one side and the other is distributed on the other side.
Therefore, surface charge density on each face of the plate is
[tex]\sigma = Q/A = \frac{2.35 \times 10^{-8}}{0.16} = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]
(b) To find the electric field just above the plate, Gauss' Law can be used. Normally, Gauss' Law can only be used in infinite sheet (considering the flat surfaces), but just above the surface can be considered that the distance from the surface is much much smaller than the length of the plate (x << l).
In order to apply Gauss' Law, we have to draw an imaginary cylinder with radius r. The cylinder has to stay perpendicular to the plane.
[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r^2 = \frac{\pi r^2 \sigma}{\epsilon_0}\\E = \frac{\sigma}{2\epsilon_0}\\E = \frac{1.46 \times 10^{-7}}{2\times 8.8\times 10^{-12}} = 8.34\times 10^3~{\rm N/C}[/tex]
The direction of the electric field is in the upwards direction.
(c) The magnitude of the electric field is the same as that of upper side. Only the direction is reversed, downward direction.
