application of gauss law

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application of gauss law


Ok hello friends now we have come to the next level of gauss law which is application of gauss law . in this article we all learned so many application of gauss law , i suggest you that if you not know about gauss law in easy language so please first visited this page which is  what is gauss law ? so that you might be introduced gauss law vary well.


So now let’s start with basic examples and then go to the next level as iit's level ok.

application of gauss law      


As we also read in the next page , the electric field due to the semicircle, so let's use that formula to solve some most important basic question.


Our first question is :


If we take a semicircle and provide a negative charge density on it then charge density will flow over semicircle and the full semicircle will be charged with 位 so find an electric field due to it?


Ans: 
While solving this question when negative charge on semicircle, keep some point in mind:


  • Electric field due to negative charge particles towards the charged body so in this case electric field will be towards semicircle and electric field due to positive charge always away from the charge .

  • Electric field due to a negatively charged body is the same as an electric field due to a positively charged body.

application of gauss law

So electric field due to negative semicircle is:

E = 2k饾潃 / R





Now we are going to solve question on gauss law :


{application of Gauss law applicable on regular closed surface bodies like sphere, cone , square, cylinder and so on , not an open body like circle, plane , disc, annular and so on.}









As we have discussed electric flux or number of electric field lines passing through one surface of cube when a charge is placed in the center of the cube if not known so check this page  electric flux . .


Q.  Now let's know what is electric flux or number of electric field lines passing through a cube plane which is made by combining four cube.

application of gauss law


Ans: 

By seeing image if we talk about number of electric field line passing through all cube surface then by gauss law :

饾瀺E = Q Enclosed / 饾瀶

{this is number of field lines passing through hole cube}



So let’s separate out one by one faces to calculate flux :

When we separate half part of the cube then the electric field ?


application of gauss law

Now when we remove half of the cube then charge is placed on the vertex of two cube and now if we see carefully then we find that electric fields line basically passing only four surfaces 1,2,3 and 4 of each cube.


So number of electric field lines passing through only two surface of both cube will be:

饾瀺E = (Q Enclosed / 饾瀶)*(1/4)

饾瀺E = Qinclosed / 4饾瀶






So now let’s remove one cube more to calculate electric flux passing through only one surface of all cube (due to if we find electric flux passing through one surface of all cube , it means that flux passing through individual surfaces of the cube also same.)


Now when we remove all cubes except only one , electric field lines passing only three face of cube 1, 2 and 3:

application of gauss law


So electric flux or number of electric field lines passing only one face of cube is:

饾瀺E = Qinclosed / 8饾瀶


  • Let's assume charge is pleasing on the edge of two cube or center of four cube so number of faces from which electric field lines pass is 8 so by unitary method electric field lines passing through only one face is 1/8 times of electric field lines that is the electric flux.

饾瀺E +饾瀺E +饾瀺E + 饾瀺E + 饾瀺E + 饾瀺E +饾瀺E + 饾瀺E = Qinclosed / 饾瀶

饾瀺E = Qinclosed / 8饾瀶



Lets understand it more clearway 馃憤


Now we place a charge on the edge of plane surface and let’s find out electric flux passing through plane surface is :

application of gauss law


So in this case if we observe that all electric field lines pass parallel to the plane surface , there are no electric field lines passing through the surface.

So due to angle between electric field vector and area vector is 90 so

    饾瀺E = EAcos(90)
                                    {as we know cos 90 = 0}

So 饾瀺E = EA* 0

    饾瀺E = 0


So flux passing through the plane surface in this case is zero.





Let’s back our topic


So when charge is pleasing on the edge of the cube then three planes are there (above case) from which there are no electric field lines passing (a , b ,c) and only three planes are there from which electric field lines are passing through it (1 ,2 ,3).

application of gauss law

So electric flux passing through only one plane is the same as the remaining two planes.

So 饾瀺E + 饾瀺E + 饾瀺E = Qinclosed / 饾瀶

so                3饾瀺E = Qinclosed / 饾瀶

                      饾瀺E = Qinclosed /3 饾瀶


  • Now I think you understand the concepts very well. If you are facing any problem then read again this article from we have started the cube problem and am sure this time you will understand very well.


Now let’s take cone problem:


Electric flux of cone (when charge pleased in center of cone)



  • As we will see in this case charge is placed in the center of the cone so first let’s learn what is electric flux passing through the curved part of the cone.

  • So if we take any ring and pass electric field lines through it from distance h the number of electric field line passing is 2Q / 5饾瀶.
    application of gauss law


  • So number of electric field lines passing through  curved surface is 3Q / 7饾瀶 (given).
application of gauss law

And total electric flux passing through the cone Q / 饾瀶 .


So let’s find electric flux passing through a curved surface.

= total flux - flux due to plane

= (Q / 饾瀶) - (3Q / 7饾瀶 )

= 4Q / 7饾瀶


So 饾瀺E( curved) = 3Qinclosed /7 饾瀶

And 饾瀺E(plane) = 4Qinclosed /7 饾瀶





let‘s take linear charge not point charge and then find out electric flux due to it.

Charge in side body



Linear charger :


Cylinder:



  • Let’s take a cylinder and placed a linear charged rod which is passing through center of cylinder and charge density of rod is lambda (位) .( if you want to learn more about charge density and its application so please visit this page also )
application of gauss law


  • So due to the linear charged rod placed inside the cylinder all electric field lines pass through curved surface not plane surface due to the area vector of the plane surface is perpendicular to the electric field line and we know cos 90 = 0 .


So electric field passing through curved surface or cylinder is :

∮E . dA = 桅net = Qin / 蓻

                         = Qin / 蓻

                                        { Qin = 位L }

                         = 位L / 蓻


∮E . dA = 桅net = 位L / 蓻
So number of electric field line passing through cylinder is 位L /蓻.





Cone:


  • Now in this case a linear charged rod is placed into the cone and it is passing through the center of the cone.


  • This is the same case as we read above: electric flux passing through a plane is zero and all electric flux will be passes through a curved surface.
    application of gauss law

  • So electric field passing through curved surface or cylinder is:


∮E . dA = 桅net = Qin / 蓻

                      = Qin / 蓻

                                     { Qin = 位( area of curved face ) }

                                           {area of curved surface is √ (L^2 - R^2)}


                       = 位[√ (L^2 - R^2)] /蓻


∮E . dA = 桅net = 位 [√ (L^2 - R^2)] /蓻



So............................number of electric field line passing through cone is 位 [√ (L^2 - R^2)] / 蓻.





Cube :


  • Now we take a cube and a linear charge in diagonal of it which is charge density 位 (lambda).
application of gauss law

  • So electric field passing through cube is:


∮E . dA = 桅net = Qin / 蓻

                      = Qin / 蓻


                        { Qin = 位( length of diagonal ) }

                       {area of curved surface is √3 a}


                       = 位[√3 a] / 蓻


∮E . dA = 桅net = 位 [√3 a] / 蓻


So number of electric field line passing through cone is 位 [√3 a] / 蓻.





Circle (my favorite )


Now we take a circle which equation [ X^2 +y^2 + Z^2 = 4^2] and pleased three linear charged rod in which two are positive charge and one is negative charged rod and all linear charged rod are passing through center of circle.

application of gauss law



Charge density of first rod = 2位

Charge density of second rod = 位

Charge density of third rod = - 位


  • So by equation of circle [ X^2 +y^2 + Z^2 = 4^2]
  • So 4^2 represent r^2 (radius square )
  • So r = 4


  • As we can see in figure linear charge rod passing through diameter of circle (D = 2r)

So enclosed charge due to first linear charge = 2位 * D

= 2位 (2 * 4)

= 8 * (2位)


Similarly enclosed charge due to second linear charge = 位 * D

= 位 (2 * 4)

= 8 * (位)


enclosed charge due to third linear charge = -位 * D

= -位 (2 * 4)

= 8 * (-位)



So number of electric field lines passing through all circle due to these linear charges:



桅net = Qin / 蓻


桅net = [8 * (2位) + 8 * (位) + 8 * (-位) ] / 蓻


桅net = [16位 ] / 蓻


So number of electric field line passing through circle is [16位 ] /蓻.







Electric flux (when surface charge density placed)

Coming soon……………



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