Electric charges and fields

 

Electric Charges and Fields

      Topic:-
15-    Application of Gauss's law.

16-   Problems

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                     Application of Gauss's law.

Electric field due to a uniformly charged infinite plane sheet

As shown in the figure, consider a thin infinite plane sheet of charge with uniform surface charge density . We wish to calculate its electric field at a point P at distance r from it.




                                Gaussian surface for a uniformly charged infinite plane sheet.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have the same magnitude and opposite direction at two points P and Pi equidistant from the sheet and on opposite sides. We choose cylindrically The gaussian surface of cross-sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder the flux through the curved surface is zero. The flux through the plane end faces of the cylinder is

 

Clearly, E is independent of r, the distance from the plane sheet.

(1)  If the sheet is positively charged (σ>0), the field is directed away from it.

(2)  If the sheet is negatively charged (σ<0), the field is directed towards it.

For a finite large planer sheet, the above formula will be approximately valid in the middle region of the sheet, away from its edges.


Electric field of two positively charged parallel plates.

In given figure shows two thin plane sheets of charges having a uniform charge density  and  with . Suppose  is a unit vector pointing from left to right.


                                  Two positively charged parallel plates

In the region 1:

Field due to the two sheets are

  


From the principle of superposition the total electric field at any point of the region I is

 

In region 2:

Field due to the two sheets are


In the region 3:

Field due to the two sheets are





  In the region 1:

Field due to the two sheets is

                                             

      


In region 2:

Field due to the two sheets are

                                              



 

 

FIELD DUE TO A UNIFORMLY CHARGED THIN SPHERICAL SHELL

Consider a thin spherical shell of charge of radius R with uniform surface charge density σ from symmetry, we see that the electric field at any point is radial and has the same magnitude at points equidistant from the centre of the shell i.e., the field is spherically symmetric. To determine the electric field at any point p at a distance r from O, we choose a concentric sphere of radius r as the Gaussian surface.



                        Gaussian surface for outside point of a thin spherical shell of charge.

 

(a) When point P lies outside the spherical shell. The total charge q inside the Gaussian surface is




This field is the same as that produced by a charge q placed at the centre O. Hence for a point outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at the centre.

(b) When point p lies on the spherical shell. The Gaussian surface just enclosed the charged spherical shell.

Applying gauss’s theorem,




(c) When point p lies inside the spherical shell. As is clear from the given figure, (a) the charge enclosed by the Gaussian surface is zero i.e.,

                                                   q = 0


             (a)   Gaussian surface for inside points of a thin spherical shell of charge.

Flux through the Gaussian surface,



Hence electric field due to a uniformly charge spherical shell is zero at all point inside the shell.

In given figure (b) shows how E varies with distance r from the centre of the shell of radius R, E is zero from r= o to r=R and beyond r= R, we have,



                                     (b)  Variation of E with r for a spherical shell of charge.




 







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