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12th Class Physics Electrostatics & Capacitance Question Bank MCQ - Electrostatic Potential and Capacitance

  • question_answer
    The distance between \[{{H}^{+}}\] and \[C{{l}^{-}}\]ions in HCl molecules is\[1.38\,\overset{{}^\circ }{\mathop{A}}\,\]. The potential due to this dipole at a distance of \[10\,\overset{{}^\circ }{\mathop{A}}\,\]on the axis of dipole is:

    A) 2.1V

    B) 1.8V      

    C) 0.2V                            

    D) \[1.2\,V\]

    Correct Answer: C

    Solution :

    (c) 0.2V Here, Charge, \[2a=1.38\times {{10}^{-10}}m,\,r=10\times {{10}^{-10}}m\] As potential, \[V=\frac{p}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}=\frac{q\left( 2a \right)}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times 1.38\times {{10}^{-10}}}{{{\left( 10\times {{10}^{-10}} \right)}^{2}}}=0.2\,V\]


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