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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by \[\rho (r)=kr,\]where r is the distance from the centre. Two charges A and B, of -Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then: [JEE Main 12-4-2019 Afternoon]

    A) \[a=\frac{3R}{{{2}^{{}^{1}/{}_{4}}}}\]                     

    B) \[a=R/\sqrt{3}\]

    C) \[a={{8}^{-1/4}}R\]

    D)   \[a={{2}^{-1/4}}R\]

    Correct Answer: C

    Solution :

    \[E4\pi {{a}^{2}}=\frac{\int_{0}^{a}{kr4\pi {{r}^{2}}dr}}{{{\varepsilon }_{0}}}\]           \[E=\frac{k4\pi {{a}^{4}}}{4\times 4\pi {{\varepsilon }_{0}}}\] \[2Q=\int_{0}^{R}{kr4\pi {{r}^{2}}dr}\]           \[k=\frac{2Q}{\pi {{R}^{4}}}\]           \[QE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{(2a)}^{2}}}\] \[R=a{{8}^{1/4}}\]


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