Page 3 - English Class X_cbse new (FINAL).cdr
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TOPIC - 1

                                                                      Coulomb’s law

                                                                      Sub Topic: 1.1

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

 1. (a)                                                                    1.                                                                                                             [1]

                                                                      [1] cont.                    2Kq2  2               −4Kq2  2                                                      [1]
                                                                                                   l 2                   l 2                                                        [3]
                                                                                  Fnet =                            +                       +  2        2Kq2  −4Kq2     cos120º
                                                                                                                                                       l2     l2

                                                                                                   Kq2  2                                 Kq2         2
                                                                                                   l 2                                  l2      
                                                                                              =                  (  4  +   16)     +  8

                                                                                                    Kq2     2
                                                                                                    l2  
                                                                                              =                  (20   +       8)

                                                                      [1]                     =  Kq2    28
                                                                                                  l2

(b)	     Potential  energy of  the system       =  4KQq + KQ2 +  Kq2  [1]                     =2     7  Kq2
                                                     a 2a         2a                                     l2

                                                                      [3]
                                                                                        (b)	

                                            OR                                                        U=        1             qA i qB       +  qBqC         +  qCqA  
(a)	                                                                                                          4πε0             AB             BC               CA   

                                                                                              PE initial =    −  −8Kq2             −  4Kq2         +   2Kq2     =  −10Kq2
                                                                                                                    l                   l                l             l

                                                                      [1] PEfinal = 0                                                              10Kq2
                                                                                                                                                      l
                                                                                  Workdone              =     PEfinal − PEinitial               =

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