2. Solved Papers
  3. MGIMS WARDHA

MGIMS WARDHA MGIMS WARDHA Solved Paper-2003

  • question_answer
    Two charges each of equal magnitude \[3.2\times {{10}^{-19}}\]coulomb but of opposite sign form an electric dipole. The distance between the two charges is 2.4 Å. If the dipole is placed in an electric field of \[5\times {{10}^{5}}\]volt/metre, then in equilibrium its potential energy will be :

    A) \[3\times {{15}^{-23}}\]joule                     

    B) \[-3.84\times {{10}^{-23}}\]joule

    C) \[-6\times {{W}^{23}}\]joule                     

    D) \[-2\times {{10}^{-}}^{26}\]joule

    Correct Answer: B

    Solution :

    For an electric dipole to be in stable equilibrium, we have Potential energy \[U=-pE\]                                 \[=-2qlE\]                                 \[=-q\times 2l\times E\] Here:\[q=3.2\times {{10}^{-19}}\]coulomb, \[2l=2.4{\AA}\]                 \[=2.4\times {{10}^{-10}}m\]                 \[E=5\times {{10}^{5}}volt/metre\] \[\therefore \]\[U=-3.2\times {{10}^{-19}}\times 2.4\times {{10}^{-10}}\times 5\times {{10}^{5}}J\]                 \[=-3.84\times {{10}^{-23}}J\]


You need to login to perform this action.
You will be redirected in 3 sec spinner