VDU_Physics_3_003

Solved University-Physics Problems

Physics 3:  Introduction to Electricity & Magnetism 

<<<  Flux of Electric Field  >>>

Javier Montenegro Joo

jmj@VirtualDynamicsSoft.com

Return to Electricity & Magnetism index page

Return to   VirtualDynamics University     home page

 

 
<<<  Flux of Electric Field  >>>
 

 (1) A wire has been bent to shape a rectangle of sides 'a' and 'b', and then it has been placed in a region where there is an electric field Eo at an angle of  75° with the x-axis. Calculate the flux of the electric field through this rectangle, when its normal makes the following angles with the Eo field. (a) 60° (b) 180° (c) 0° (d) 90°

Solution.-

The 75° orientation of the electric field plays no role in the solution of this problem because the angles between the normal to the rectangle and the electric field are data.

[01234]

 

(2) An electric field Eo is oriented so that it makes an angle of 60° with the y-axis. A disk of diameter D is placed in the region where the Eo is set up. Determine the flux of the electric field through the disk, when between its normal and the x-axis there is an angle of:      (a) 30°    (b) 90°     (c) 120°    (d) 0°

Solution.

The angle between the electric field and the y-axis is 60°.   Let “A” be the angle between the normal to the disk and the x-axis:

 

(3) A uniform electric field given by    E = 7x + 3y + 9z     has been established in a region where a cube of side L has been placed on the positive sides of the x, y, and z coordinated axes. Determine the flux of electric field through: (a) The face of the cube perpendicular to x-axis (b) The face of the cube perpendicular to y-axis (c) The face of the cube perpendicular to z-axis. (d) The total flux through the six faces of the cube.

Solution.-

The figure shows the cube placed in the electric field and shows also the case (a), when the normal to the face of the cube coincides with the x-axis and, (b) when the normal to the face of the cube is along the y-axis:

[56789]

(4) A compact ball of radius R, made of rubber, has a volume charge density that varies with the radius according to  D = Do r,  where Do  is a constant.  (a) Determine the flux of electric field through a hemispherical shell of radius 2R, surrounding the ball (b) repeat the problem for the case when the radius of the shell is  7R / 2.

 Solution.-

  In order to compute the flux through the surface of the hemisphere, the total charge inside the hemisphere must be determined. Next the electric field ‘E’  at the surface of the hemisphere must be found, finally the flux may be computed.  In part (b) the problem is repeated only to show that no matter the size of the hemisphere around the charge, the flux is the same.  In both cases the total charge inside the hemisphere is the same.

Physics Virtual Lab, PVL:    Optics simulation module:

 The Physics Virtual Lab simulates a set of several refraction plates, one on top of another, whose individual indexes of refraction can be controlled by the user.   Download a demo:

Download  Physics Virtual Lab (PVL)  Demo

 

[5] The flux of electric field through one of the faces of a cubical box of side L is Fo .  (a) Determine the total charge Qo inside the box (b) Once the total charge Qo is known, it is placed inside a sphere of radius R, find the flux of electric field through the sphere.

Solution.-

The data mentions the flux through a single face of the cube, to find the total charge inside the cube, the total flux through its 6 faces must be considered.       According to Gauss’ theorem the flux of electric field through a closed surface containing a charge Q in its interior is given by

Return to Electricity & Magnetism index page