GAUSS'S LAW & ITS APPLICATIONS

GAUSS'S LAW:-

Gauss's law is used to find total electric flux and total electric field due to number of charges enclosed by the surface

EXPLANATION & DERIVATION:-

                  Consider a closed surface enclosing number of charges q_1, q_2, q_3,--------- q_n randomly distributed inside the surface then electric flux due to charge q_{1 is}

_{USES OF GAUSS'S LAW:-}

               _{It is used to find the value of electric field and electric flux due to charged bodies enclosed in a surface}

_{LIMITATIONS OF GAUSS'S LAW:-}

              _{Gauss's law can not be used on an open surface. Any point P where electric field and electric flux is to be determined is taken on an imaginary closed surface known as Gaussian surface. If this closed surface is not given then we draw it ourselves.}

_{APPLICATIONS OF GAUSS'S LAW:-}

• INTENSITY OF ELECTRIC FIELD INSIDE A CHARGED SPHERE:-

 Consider a hollow conducting sphere of radius "a". Its outer surface is uniformly positively charged such that there is no charge at its centre i.e q=0. If we consider any point P inside the sphere where both electrical field and electric flux are to be determined then we will have to apply Gauss's law. But this law is only valid on closed surface and not on open surface so we will draw an imaginary Gaussian surface of radius "r" such that a>r inside the sphere and point P lies on this Gaussian surface. Then by gauss's law

      

_CONCLUSION:-

_{Interior of a hollow charged sphere is a field free region.}

• ELECTRIC FIELD INTENSITY & ELECTRIC FLUX DUE TO AN INFINITE SHEET:-

       Consider a uniformly positively charged infinite sheet whose area is "A" and magnitude of charge "q"

then surface charge density
surface charge density =      -------(i)
A finite part of this infinite sheet is shown in the fig. We take finite part of infinite sheet because since the charge is uniformly divided so electric field is same all along the sheet. If we consider any point P close to this charge sheet where both electric field and electric flux are to be determine then we will have to apply Gauss's law but this law is applied on closed Gaussian surface so we draw this Gaussian surface in the form of cylindrical rod as shown in the fig.We have drawn rod like Gaussian surface because electric field intensity and electric flux at point P is only due to few positive charges so Gaussian surface make that charges closed along with point P as shown in the fig.
Electric field for outer curve surface is 

_CONCLUSION:-

               _{Electric field intensity is directly proportional to surface charge density or the ratio of q and A.}

• ELECTRIC FIELD AND ELECTRIC FLUX BETWEEN OPPOSITELY CHARGED PARALLEL PLATES:-

Consider two oppositely charged parallel plates in which electric field is directed from positive plate to negative plate if we consider any point P inside the electrical field where both electric field and electric flux are to be determined then we will have to apply Gauss's law but this law is only applicable on closed surface so we draw an imaginary Gaussian surface. In the present case we draw this surface in the shape of rectangle whose three sides lies inside the electric field but fourth side lies inside positive plate where F=0 as shown in fig. Each plate has same magnitude of charge "q" and same surface area "A"so that each plate has surface charge density 

  

FOR SIDE AB:-

ϑ=0 degree between 

&

  Φ_{E =} E A cosϑ=EA cos0


         Φ_{E =EA------(iv)}

FOR SIDE BC:-

            ϑ=90 degrees  between E and A

so  Φ_{E =} E A cosϑ=EA cos90

 Φ_{E =0----(v)}

FOR SIDE CD:-

  ϑ=180 degrees between E and A but F=0 so consequently E=0 so

Φ_{E =0-------(vi)}

_{FOR SIDE DA:-}

  ϑ=270 degrees   between E and A

so   Φ_{E =} E A cosϑ=EA cos270

 Φ_{E =0----(vii)}

_{Thus the total electric flux of whole rectangular surface is} 

 Φ_{E =EA+0+0+0 of}

 Φ_{E =EA----(viii)}

_{put (viii) in (iii) we get}

CONCLUSION:-

 Electric field intensity is directly proportional to surface charge density.