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Physics 3:  Introduction to Electricity & Magnetism 

<<<  Gauss Theorem of Electric Field  >>>

Javier Montenegro Joo

jmj@VirtualDynamics.Org

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<<<  Gauss Theorem of Electric Field  >>>

(1) A compact metallic sphere of radius  r1 , having an electric charge Qo, is placed in the center of a metallic spherical hull of inner radius r2  and outer radius r3 .  Determine the electric field in:        (a) r < r1          (b)    r1 < r < r2             (c)     r2 < r < r3          (d) r > r3

Solution.-

This problem deals with conductive materials and, it is known that in these materials the electric charge distributes on the surface, hence it doesnt 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 < r1

Since the central sphere of radius r1  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 r1, having an electric charge  Qo /2  is in the center of a metallic spherical hull   of radiuses r2  and r3 , with r2 < r3, which is loaded with a charge 3Qo.  Determine the electric field in:  (a) r < r1   (b) r1 < r < r2    (c) r2 < r < r3    (d) r > r3

Solution.-

This problem deals with conductive materials and, it is known that in these materials the electric charge distributes on the surface, hence it doesnt 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 < r1

Since the central sphere of radius r1  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)  r1 < r < r2    

 In this case the charge enclosed by a Gaussian spherical surface of radius r between r1 and r2 is     -Qo/2, then

 (c) (c)  r2 < r < r3    

The charge Qo/2  at the inner piece induces a charge  Qind-2 = +Qo/2   on the surface of radius r2, hence the total charge inside the Gaussian surface with radius between r2 and r3  is zero, and E = 0.

(b)   (d)  r > r3

The induced charge  Qind_2 = Qo/2  on the inner surface of radius r2   creates an additional induced charge  Qind_3 = -Qo/2   on the external surface of radius r3 .  In these conditions the total charge enclosed by a spherical Gaussian surface with radius r > r3  is: 

 

 
(4) A copper-made 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

 

 (5) A hollow ball made of plastic material, with radiuses interior a and exterior b  has a non-uniform (volumetric) density of electric charge D = Do  / r,     Do  = 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

Algorithmic Art images by Javier Montenegro Joo.

JMJ has devised mathematical algorithms to generate images with interesting and astonishing results.

(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.

 

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