The electric flux through the closed surface in the following figure is: positiv

Question

The electric flux through the closed surface in the following figure is: positive. negative. zero. Impossible to determine from the information given. Find the electric flux through the surface in the figure below: 3C/0 6C/Two spherical shells have a center. A -2.0 mu C charge is spread uniformly over the shell, which has a radius of 0.050 m. A 6.0 mu C charge uniformly over the outer shell, which has a radius of 0.15 m. Find the magnitude and of the electric field at a distance(measured from the)of 0.20 m, 0.10 m, and 0.025 m.

Explanation / Answer

electric flux = 0 <<<====ANSWER

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(2)

from gass law

flux = Qinside/e0

flux = (-5 + 2 - 5)/(8.85*10^-12) = -0.91*10^12 Nm^2/C <<<====ANSWER

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(3)

Qinner = -2 uC

Qouter = 6 uC

Rinner = 0.05 m

Router = 0.15 m

(a)

at r = 0.2 > Router

from gauss law = Qinside/e0

E*4*pi*r^2 = (Qinner+Qouter)/e0

E*4*pi*0.2^2 = (-2+6)*10^-6/(8.85*10^-12)

E = 9*10^5 N/C         <<<====ANSWER

(b)

r = 0.1 m

Router > r > Rinner

from gauss law = Qinside/e0

E*4*pi*r^2 = (Qinner)/e0

E*4*pi*0.2^2 = (2)*10^-6/(8.85*10^-12)

E = 4.5*10^5 N/C      <<<====ANSWER

(c)

r = 0.025 m

r < Rinner

from gauss law = Qinside/e0

E*4*pi*r^2 = (0)/e0

E*4*pi*0.2^2 = 0

E = 0 N/C <<<<====ANSWER

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