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

  • question_answer
    The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance \[5\times {{10}^{-11}}m,\] will be (Charge on electron \[=\text{ 1}.\text{6}\times \text{1}{{0}^{\text{19}}}C\], mass of electron \[=\text{ 9}.\text{1}\times \text{1}{{0}^{\text{31}}}kg\], mass of proton = \[1.6\times {{10}^{-27}}kg,\] \[\,G=6.7\times {{10}^{-11}}\,N{{m}^{2}}/k{{g}^{2}})\]                  [RPET 1997; Pb PMT 2003]

    A)                    \[\text{2}.\text{36}\times \text{1}{{0}^{\text{39}}}\]

    B)                    \[\text{2}.\text{36}\times \text{1}{{0}^{40}}\]           

    C)             \[\text{2}.\text{34}\times \text{1}{{0}^{41}}\]        

    D)                    \[\text{2}.\text{34}\times \text{1}{{0}^{42}}\]

    Correct Answer: A

    Solution :

       Gravitational force \[{{F}_{G}}=\frac{G{{m}_{e}}{{m}_{p}}}{{{r}^{2}}}\] \[{{F}_{G}}=\frac{6.7\times {{10}^{-11}}\times 9.1\times {{10}^{-31}}\times 1.6\times {{10}^{-27}}}{{{(5\times {{10}^{-11}})}^{2}}}\]= 3.9 ´ 10?47 N Electrostatic force \[{{F}_{e}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{{{r}^{2}}}\] \[{{F}_{e}}=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times 1.6\times {{10}^{-19}}}{{{(5\times {{10}^{-11}})}^{2}}}\] = 9.22 ´ 10?8 N So, \[\frac{{{F}_{e}}}{{{F}_{G}}}=\frac{9.22\times {{10}^{-8}}}{3.9\times {{10}^{-47}}}=2.36\times {{10}^{39}}\]


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