Gauss’ Law

Gauss’ Law

Area vector :  In many cases, it is convenient to treat area of a surface as a vector. The length of the vector represents the magnitude of the area and its direction is along the outward drawn normal to the surface.

Mathematically

“Electric flux through an elementary area is defined as the scalar product of area vector and electric field vector” i.e

Hence flux from complete area (S)

                       

                   

If θ = 0o i.e.  electric field  is perpendicular to the surface area ,flux linked with it will be max.

               

  • SI. unit of Electric flux–        (volt × m) or  N. m2 /C
  • It’s Dimensional formula  – [ML 3T-3A-1]

Note: For a closed body outward flux is taken to be positive, while inward flux is taken to be negative.

Question1: The electric field in a region is given by   with  E0 = 2.0 × 103 N/C. Find the flux of this field through a rectangular surface of area 0.2m2 parallel to the Y–Z plane.     [Ans.240Nm2/C ]

Gauss’s Law

 “Total electric flux through a closed surface is  times  to the total electric charge enclosed by the surface”   

 i.e.                                        

Gaussian Surface: 

“The imaginary closed surface chosen around a charged body to apply Gauss’s law, is called Gaussian Surface.”

            This surface should pass through the point where electric field is to be calculated and must have a shape according to the symmetry of source.

Zero flux :  The value of flux is zero in the following circumstances.

                     

 Note: If a closed body (not enclosing any charge) is placed in an electric field (either   uniform or non-uniform) total flux linked with it will be zero.

                 

Question: A point charge Q is placed at the center of the plane surface of a hemisphere. Calculate the total flux through the hemispherical body.[Ans:  ]

Question : If a point charge Q is located at the centre of a cube then find (i) flux through the total surface, (ii) flux through one surface.

Question: A point charge ‘Q’ is kept at one of the corner of a cube of side ‘a’. Find 

  • the total electric flux through the cube.
  • The flux through the sides of the cube. [ Ans (a) (b) through three surfaces and zero through remaining three]

Question: Although, while calculating the net electric field at point on the surface all the charges, inside or outside, are taken into account, the total electric flux through the surface depends only upon the charges present inside the surface. Why?

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