Page 13 - English Class X_cbse new (FINAL).cdr
P. 13

S.                                               Solution                 Marking   S.                                                        Solution                           Marking
No.                                                                        Scheme   No.                                                                                           Scheme

4. According to Gauss theorem,                                                 [1]  5. Permittivity of free space: (εo)                                                           [1]
                                                                               [1]
Cont. φ = q                                       (q = σA)                     [3]  cont.           1                                  C2
           ε0                                                                                     4πε0                               N ⋅ m2
                                                                                           k   =           =     8.85      ×  10−12

     φ = σA                                                                                Gauss’s law in electric flux φE through any closed surface is equal to
           ε0 				…(ii)                                                                    the net charge inside the surface, Q inside, divided by ε0 .

     From equations (i) and (ii)                                                           φE  =     Qinside
     2EA = σA                                                                                          ε0

               ε0

     E= σ                                                                                  (b) Let charge is situated at centre of sphere of radius r.
          2ε0
                                                                                                                                                            E

     The direction of field for positive charge density is in outward                                q                      P
     direction and perpendicular to the plane infinite sheet. Whereas                                   O           r ds
     for the negative charge density the direction becomes inward and                                               r
     perpendicular to the sheet.

5. (a)	The electric flux through a closed surface is proportional to the
           charge contained inside the surface.

                                                                                           According to coulomb’s law,

                                           qq                                              E   =    1         ⋅  q      r
                                                                                                  4π ∈0          r2

                                       S1                        S1                        where r is unit vector along the line OP consider a small area                         [1]
                                     S2                                                    element ds around the point P. Since the small area element is                           11
                                                                S2                         located on the surface of sphere, the area vector ds will also be
     	 S3                                                   S3
                                                                                           along OP i.e., in direction of unit vector r .
     The strength of the electric field everywhere on this sphere is

     E  =  k   q                                                           [1]             Electric flux through area element ds is
               r2
                                                                                                                      1          q                  1       q                (1)
     Since every point on the sphere’s surface is a distance r from the                    dφ     =  E ⋅ ds   =     4π ∈0     ⋅  r2  ⋅ r ⋅ ds  =  4π ∈0  ⋅  r2  ds  cos  0°

     charge

     ( )φE               q                                                                 dφ  =       1         ⋅  q   ⋅ ds
        =  E   A   =  k  r2                4πr 2  = 4πkq                                             4π ∈0          r2
   8   9   10   11   12   13   14   15   16   17   18