33.6k views
2 votes
A dipole is located at the origin, with charges +3 nC and -3 nCseparated by 4 μm. Consider a spherical Gaussian surfacecentered on the origin of radius 80 μm.

a) What is the electric flux through the Gaussiansurface?
b) Now suppose you have a Gaussian sphere of radius1 μm centered on the positive charge. What is the fluxthrough this Gaussian surface?

1 Answer

2 votes

Answer:

Step-by-step explanation:

a )

The Gaussian surface must have contained both the charges like +3 nC and - 3 nC , total charges being zero . According the Gauss's law , total electric flux over a closed surface is equal to 1 / ε₀ times the total charge contained inside the closed surface .

Since total charge inside is zero , total flux will be 1 / ε₀ x 0 = 0

b )

In this case the Gaussian surface is so small that it contains only the charge

+ 3 nC .

So electric flux over the Gaussian surface

= ( 1 / ε₀ ) x 3 x 10⁻⁹

= (1 / 8.85 x 10⁻¹² ) x 3 x 10⁻⁹

= 3 x 10³ / 8.85

= 339 weber.

=

User Topek
by
5.1k points