Gauss\' Law states that the electric flux through a closed \"Gaussian\" surface

Question

Gauss' Law states that the electric flux through a closed "Gaussian" surface equals the net charge inside divided by the constant

(a) Suppose that a charge -4.0 nC is located at the origin, and that the Gaussian surface is a sphere of radius 15.0 cm centered on the origin. What is the electric flux through the sphere?_____ N m2/C

(b) The Gaussian sphere is shifted so that its center is at x = 5.0 cm. What is the flux now? ____N m2/C

(c) The Gaussian sphere is shifted so that its center is at x = 35.0 cm. What is the flux now?____ N m2/C

(d) A new Gaussian sphere is used of twice the original radius centered on the origin. What is the flux now?____ N m2/C

(e) A new Gaussian surface is a tetrahedron centered on the origin with sides of length 15 cm. What is the flux now?___ N m2/C

(f) A second charge +6.0 nC is added located at x = 17.0 cm. A Gaussian sphere of radius 15.0 cm is centered on the origin. What is the flux through the sphere? ____N m2/C

(g) The sphere in part (f) is now centered at x = 5.0 cm. What is the flux through the sphere? ____ N m2/C

Explanation / Answer

a) phi_E = q_inside/eo = (4*10^-9)/(8.85*10^-12) = 452 Nm^2/C

b) at x = 5 cm

phi_E = 452 Nm^2/C

c) at x = 35 cm

since no charge is there inside the gaussinal surface then

phi_E = 0 Nm^2/C

d) increasing radius wouldn't effect the flux ,hence

phi_E = 452 Nm^2/C

e) For tetrahedron

phi_E = 452 Nm^2/C

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