Electric Field Flux Problems.

**Gauss' Law Exercises, Problem One**

**Using Gauss' Law, obtain the electric field
created by a charge uniformly distributed over a plane.**

__Solution.-__ The figure represents
a plane which carries a positive charge a per unit area. The electric field
is perpendicular to the plane due to the symmetry of the problem. Choosing
for our closed surface the cylinder shown in the figure, we may separate the
electric flux Ø into three terms: The flux through S1 which is +ES, where
S is the area of the base of the cylinder ; the flux through S_{2},
which is also +ES, and the flux through the lateral surface of the cylinder,
which is zero because the electric field is parallel to the surface. Therefore
we have Ø = 2ES. The charge inside the closed surface is that in the shaded
area and is equal to q = aS. Applying Gauss' Law we have

**Gauss' Law Exercises, Problem Two**

**A solid conductor sphere of radius a has
a positive net charge 2Q. A conductor spherical shell of inner radius b and
external radius c has the same center with the solid sphere and has a net
charge -Q. Using Gauss' Law determine the electric field in the regions denoted
as (1), (2), (3) y (4) in the figure below, and the charge distribution over
the shell when all the system is in electrostatic equilibrium.**

__Solution.-__ To find E in the inside
of the solid sphere, region (1), consider a gauss' surface of radius r < a.
Since there is no charge inside a conductor in electrostatic equilibrium,
it is clear that q_{in} = 0. Then, based on
Gauss' Law and the symmetry, E1 = 0 for r < a.

**In the region (2) - between the solid sphere
surface and the inner shell surface - consider a spherical gauss' surface
of radius r where a < r < b. The charge inside this surface is +2Q (the charge
on the solid sphere). Due to the spherical symmetry the electric field lines
must point with its radius outward and have constant magnitude over the Gauss'
surface. Using Gauss' Law it is found that**

**In the region (4), where r > c, the spherical
Gauss' surface encloses a total charge q _{in}
= 2Q - Q = Q. Then, Gauss' Law applied to this surface produces**

**In the region (3) the electric field must
be zero due the spherical shell is also a conductor in equilibrium. As E _{3}
= 0, the Gauss Law applied to the surface with radius r where b < r < c implies
q_{in} = 0. Therefore the charge over the inner
spherical shell surface must be -2Q to cancel the + 2Q charge on the solid
sphere. As the net charge on the shell is - Q, it is concluded that the external
shell surface must have a +Q charge.**

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