Page 24 - Electrostatics 11
P. 24

Grade 11                                                                              © GC Shiba


                                                   =E . A
                                                              
                                            or,  E . A =
                                                              0
                                                      =      
                                                                0
                                                                 
                                                       =
                                                           4        2
                                                                0


               Hence, charged spheres behave as whole charges are concentrated at its
        center.


        3) Electric intensity on the surface of a charged conducting

            sphere

               Consider a charge conducting sphere of radius R
        having a charge +Q. We are interested to find the
        electric intensity at point P on the surface of  a charged

        sphere. Let us draw a concentric sphere of radius r
        through point P. This sphere is called a Gaussian sphere.                                     r = R     P
        By symmetry, electric intensity would be the same at all

        points on the surface of a Gaussian sphere. Electric flux
        from the charged sphere is radially outward as shown in
        the figure.

        Now, electric flux passing through the gaussian surface
        is
                                                                        ℎ        
                                                   =
                                                               ∈ 0

                                                   =E . A

                                                              
                                            or,  E . A =
                                                              0
                                                      =      
                                                                0
                                                                 
                                                       =
                                                           4        2
                                                                0

               Hence, the electric field just outside the surface.






        24                                                                                      Electrostatics
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