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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Charge and Coulombs Law

  • question_answer
    An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force \[\overrightarrow{F}\] between the two is  (Where \[K=\frac{1}{4\pi {{\varepsilon }_{0}}}\])      [CBSE PMT 2003]

    A)            \[-K\frac{{{e}^{2}}}{{{r}^{3}}}\hat{r}\]                          

    B)            \[K\frac{{{e}^{2}}}{{{r}^{3}}}\vec{r}\]

    C)            \[-K\frac{{{e}^{2}}}{{{r}^{3}}}\vec{r}\]                         

    D)            \[K\frac{{{e}^{2}}}{{{r}^{2}}}\hat{r}\]

    Correct Answer: C

    Solution :

                 \[\overrightarrow{F\,}=-\,k\frac{{{e}^{2}}}{{{r}^{2}}}\hat{r}=-\,k.\frac{{{e}^{2}}}{{{r}^{3}}}\overrightarrow{r\,}\]                 \[\left( \because \,\,\hat{r}=\frac{\overrightarrow{r\,}}{r} \right)\]


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