JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    Two point dipoles of dipole moment \[\overset{\to }{\mathop{{{\operatorname{P}}_{1}}}}\,\] and \[\overset{\to }{\mathop{{{\operatorname{P}}_{2}}}}\,\] are at a distance \[x\] from each other and \[\overset{\to }{\mathop{{{\operatorname{P}}_{2}}}}\,||\overset{\to }{\mathop{{{\operatorname{P}}_{2}}}}\,\]. The force between the dipoles is:                   JEE Main Online Paper (Held On 09 April 2013)

    A)                 \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4{{\operatorname{p}}_{1}}{{\operatorname{p}}_{2}}}{{{x}^{4}}}\]                

    B)                 \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{3{{\operatorname{p}}_{1}}{{\operatorname{p}}_{2}}}{{{x}^{3}}}\]                               

    C)                 \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{6{{\operatorname{p}}_{1}}{{\operatorname{p}}_{2}}}{{{x}^{4}}}\]                

    D)                 \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{8{{\operatorname{p}}_{1}}{{\operatorname{p}}_{2}}}{{{x}^{4}}}\]                

    Correct Answer: B

    Solution :

                                    Both the charges of each dipole will interact with both the charges of to the dipole with an electrostatic force and then by vector sum, we can find the net force on the dipole.                 \[{{F}_{net}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{3{{p}_{1}}{{p}_{2}}}{{{X}^{4}}}\]                


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