JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    From a conducting ring of radius R which carries a charge Q (uniformly distributed) along its periphery a small length \[d\ell \] is cut off. The electric field at the  center due to the remaining wire is

    A) \[\frac{Qd\ell }{8{{\pi }^{2}}{{\in }_{0}}{{R}^{3}}}\]                          

    B) \[\frac{Q}{4\pi {{\in }_{0}}{{R}^{2}}}\]

    C)  zero                     

    D)  can't be determined

    Correct Answer: A

    Solution :

    Here to solve this quations we can use principle of superposition. The given structure can be considered as combination of two as shown in figure. \[{{\vec{E}}_{at}}\] (due to given structure) \[={{\vec{E}}_{atO}}(due\,to\,l)-{{E}_{atO}}(due\,to\,all)\]\[E=\frac{dq}{4\pi {{\in }_{0}}{{R}^{2}}}\]towards dl. \[=\frac{Q/2\pi R\times d\ell }{4\pi {{\in }_{0}}{{R}^{2}}}=\frac{Qd\ell }{8{{\pi }^{2}}{{\in }_{0}}{{R}^{3}}}\]

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