VDU_Physics_3_006 
Solved UniversityPhysics ProblemsPhysics 3: Introduction to Electricity & Magnetism <<< Gauss Theorem of Electric Field >>>Javier Montenegro Joo jmj@VirtualDynamics.OrgReturn to Electricity and Magnetism Index page Return to VirtualDynamics University home page


<<< Gauss Theorem of Electric Field >>> 
(1) A compact metallic sphere of radius r_{1 }, having an electric charge Qo, is placed in the center of a metallic spherical hull of inner radius r_{2 } and outer radius r_{3 }. Determine the electric field in: (a) r < r_{1} (b) r_{1 }< r < r_{2 } (c) r_{2 }< r < r_{3 } (d) r > r_{3} Solution. This problem deals with conductive materials and, it is known that in these materials the electric charge distributes on the surface, hence it doesn’t matter whether the inner metallic sphere is compact or just a skin. Additionally there are induced charges on both surfaces of the outer hull which must be taken into account in order to compute the electric field. (a) r < r_{1} Since the central sphere of radius r_{1 } is metallic, the charge Qo distributes on its surface, existing no charge inside this central object, consequently the electric field inside this piece is zero: E = 0.


(2) A
metallic sphere of radius r_{1},_{ }having an electric
charge –Q_{o }/2 is in the center of a metallic spherical hull
of radiuses r_{2 } and r_{3 }, with r_{2} < r_{3},
which is loaded with a charge 3Q_{o}. Determine the electric
field in: (a) r < r_{1} (b) r_{1 }< r < r_{2 }
(c) r_{2 }< r < r_{3 } (d) r > r_{3} Solution.
This problem deals with conductive materials and, it is known that in these materials the electric charge distributes on the surface, hence it doesn’t matter whether the inner metallic sphere is compact or just a thin casing. Additionally there are induced charges on both surfaces of the outer hull which must be taken into account in order to compute the electric field. (a) (a) r < r_{1} Since the central sphere of radius r_{1 } is metallic, the charge –Qo/2 is distributed on its surface, existing no charge inside this sphere, consequently the field inside this piece is zero: E = 0. (b) (b) r_{1 }< r < r_{2 } In this case the charge enclosed by a Gaussian spherical surface of radius r between r_{1} and r_{2} is Qo/2, then
(c) (c) r_{2 }< r < r_{3 } The charge –Qo/2 at the inner piece induces a charge Q_{ind2 }= +Qo/2 on the surface of radius r_{2}, hence the total charge inside the Gaussian surface with radius between r_{2 }and r_{3 } is zero, and E = 0. (b) (d) r > r_{3} The induced charge Q_{ind_2 }= Qo/2 on the inner surface of radius r_{2 } creates an additional induced charge Q_{ind_3} = Qo/2 on the external surface of radius r_{3 }. In these conditions the total charge enclosed by a spherical Gaussian surface with radius r > r_{3} is:

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(4) A coppermade spherical shell whose exterior radius is R = 4 cm, has
an inner radius whose length is half of its outer radius. The shell is
electrically charged and, at an exterior distance r it produces an
electric field given by
Calculate the total electric charge in the shell. Solution

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(5) A hollow ball made of plastic material, with radiuses interior ‘a’ and exterior ‘b’ has a nonuniform (volumetric) density of electric charge D = D_{o }/ r, D_{o } = Constant. Calculate the electric field at a radial distance measured from the center of the ball. (a) exterior (r > b ) (b) interior ( r < a ) Solution

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(6) A spherical distribution of electric charge of radius R has a charge density (a) Determine the total charge in the sphere (b) the electric field at a radius r <= R Solution.

(7) A conductive sphere of radius b, produces an electric field
at an exterior distance r from the sphere. Determine (a) the total charge in the sphere (b) The charge density in the sphere. Solution.

(8)
A
long metallic bar has an electric charge density Lambda. Calculate the
electric field produced by this charge (a) Outside the bar at a radial
distance r (b) Inside the bar at a radial distance r. Solution In any conductive material –like a metal the electric field is necessarily radial. (a) Consider a Gaussian cylindrical surface of length L and radius r around the bar and apply the Gauss Law of electric field. The direction of the electric field at this Gaussian surface is radial and so is that of the normal vector to the surface. In this way
(b) Since the bar is metallic all the charge is on its surface, thus there is no electric charge in the interior of the bar, hence the electric field there is zero. 
(9)
A very long cylinder
of radius R has a volumetric electric charge density given by
For a cylinder segment of length L and far from the extremes, determine: (a) The total charge (b) the external electric field at a distance r >= R (c) the internal electric field at a distance r <= R. Solution Since the evaluation of the electric field is considered far from the cylinder extremes, the resulting electric field is radial. It is worthwhile mentioning at this point that at the extremes of the cylinder the electric field is not radial, it is rather complicated. Another important point to bear in mind is that the internal electric field at r <= R is produced by a fraction of the total charge in the cylinder.

[10]
In the figure, a metallic spherical
shell of inner radius ‘a’ and outer radius ‘b’, has a point charge Q_{o
} in its center. It is known that the total charge in the shell is
4Q_{o }. Determine: (a) The charge densities at the surfaces
with radiuses ‘a’ and ‘b’ (b) the external electric field (c) the
electric field for r < a (d) The electric field between the inner and
the outer surfaces.
Solution. Since the shell is metallic all the electric charge is on its surface. The central point charge Q_{o } induces a surface charge –Q_{o }at the inner surface so that the charge density at this surface is:

[11] A spherical distribution of charge of radius R has an electric charge density
Determine the electric field for: (a) r > R (b) r < R Solution. In order to compute the exterior electric field the total charge in the sphere must be calculated, next the Gauss theorem must be applied. On the other hand, to compute the inner (inside the sphere) electric field, the corresponding fraction of the total charge must be found before applying the Gauss theorem.

[12] A very long cylindrical distribution of electric charge has a radius R and a charge density that varies with the radius according to
Determine the radial electric field at: (a) r > R (b) r < R Solution. [a] In order to compute the exterior electric field the total charge in a length L of the cylinder must be calculated, next the Gauss theorem must be applied to a cylindrical Gaussian surface surrounding the cylinder. [b] To compute the inner (inside the cylinder) electric field, the corresponding fraction of the total charge must be found before applying the Gauss theorem to a Gaussian cylindrical surface inside the cylinder.
