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
where C2/(N m2).
(a) Suppose that a charge -3.0 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 +4.5 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) For a closed Gaussian surface, the electric flux is simply: = q/

where:

= electric flux,

q = charge within the Gaussian surface

= permitivitty of free space = 8.85 x 10-12  m-3 kg-1 s4 A2

Thus, the Electric flux through the sphere is just = (-3.0 nC) / ( 8.85 x 10-12 ) = -339  kg m3 s3 A1

b) When the centre of the sphere is now at 5.0 cm, since its radius is 15.0 cm, the charge is still within the Gaussian surface. Since positioning of the charge does not affect the amount of electric flux through the Gaussian surface, the flux through the sphere is still -339  kg m3 s3 A1

c) When the centre of the sphere is now at x = 35.0 cm, there is now no net charge within the sphere (i.e. the Gaussian surface). Thus the flux through the sphere is now zero.

d) The electric flux is independent of the radius of the sphere, since the field lines passing through the Gaussian surface remains constant, since the amount of charge within the Gaussian surface remains constant. Thus the electric flux is the answer in part A: -339  kg m3 s3 A1

e) Electric flux is also independent of shape of the Gaussian surface, as long as it is closed surface. Thus the electric flux is -339  kg m3 s3 A1

f) Any external charges does not change the electric flux of the Gaussian surface. We only take in to account the charges within the Gaussian surface, which in this case is the charge of -3.0 nC. Thus the electric flux is still -339  kg m3 s3 A1

g) Ok now the new charge is within the Gaussian surface. So now we have to take into account its effects. Thus, the net charge in the sphere is now : -3.0 nC + 4.5 nC = 1.5 nC. This will now give rise to a new electric flux as follows:

= q/ = (1.5 nC) / ( 8.85 x 10-12 ) = 169 kg m3 s3 A1

Therefore, the new electric flux after the new charge was added into the Gaussian surface is:

= 169 kg m3 s3 A1

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