NEET Physics Electrostatics & Capacitance Question Bank NEET PYQ - Electrostatics and Capacitance

  • question_answer
    Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron)   [AIPMT (S) 2010]

    A) \[\frac{4\pi {{\varepsilon }_{0}}F{{d}^{2}}}{{{e}^{2}}}\]

    B) \[\sqrt{\frac{4\pi {{\varepsilon }_{0}}F{{e}^{2}}}{{{d}^{2}}}}\]

    C) \[\sqrt{\frac{4\pi {{\varepsilon }_{0}}F{{d}^{2}}}{{{e}^{2}}}}\]

    D) \[\frac{4\pi {{\varepsilon }_{0}}F{{d}^{2}}}{{{q}^{2}}}\]

    Correct Answer: C

    Solution :

    [c] Two positive ions each carrying a charge q are kept at a distance d, then it is found that force of repulsion between them is                       
                \[F=\frac{kqq}{{{d}^{2}}}\]
                \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qq}{{{d}^{2}}}\]
                where    \[q=ne\]
    \[\therefore \]      \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{n}^{2}}{{e}^{2}}}{{{d}^{2}}}\]
                \[\Rightarrow \]   \[n=\sqrt{\frac{4\pi {{\varepsilon }_{0}}F{{d}_{2}}}{{{e}^{2}}}}\]


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