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

  • question_answer
    \[{{F}_{g}}\] and \[{{F}_{e}}\] represents gravitational and electrostatic force respectively between electrons situated at a distance 10 cm. The ratio of \[{{F}_{g}}/{{F}_{e}}\] is of the order of                [NCERT 1978; CPMT 1978]

    A)                                                      \[{{10}^{42}}\]                     

    B)                    10

    C)                    1                                        

    D)                    \[{{10}^{-43}}\]

    Correct Answer: D

    Solution :

                       Gravitational force between electrons \[{{F}_{G}}=\frac{G{{({{m}_{e}})}^{2}}}{{{r}^{2}}}\] Electrostatics force between electrons \[{{F}_{e}}=k.\frac{{{e}^{2}}}{{{r}^{2}}}\] \[\frac{{{F}_{G}}}{{{F}_{e}}}=\frac{G{{({{m}_{e}})}^{2}}}{k.{{e}^{2}}}=\frac{6.67\times {{10}^{-11}}\times {{(9.1\times {{10}^{-31}})}^{2}}}{9\times {{10}^{9}}\times {{(1.6\times {{10}^{-19}})}^{2}}}\]\[=2.39\times {{10}^{-43}}\]


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