12th Class Physics Electric Charges and Fields Question Bank MCQ - Electric Charges and Fields

  • question_answer
    Five charges \[{{q}_{1}},\,{{q}_{2}}\]\[{{q}_{3}},{{q}_{4}}\] and \[{{q}_{5}}\]are fixed at their positions as shown in figure, S is a Gaussian surface. The Gauss's law is given by \[\oint{_{s}}\overrightarrow{E}\,.\,d\,\overrightarrow{S}=\frac{q}{{{\varepsilon }_{0}}}\] Which of the following statement is correct?                                                            (NCERT EXEMPLAR)

    A) \[\overrightarrow{E}\]on the LHS of the above equation will have a contribution from \[{{q}_{1}}.\,{{q}_{5}}\]and \[{{q}_{3}}\]while q on the RH5 will have a contribution from \[{{q}_{1}}\]and \[{{q}_{4}}\] only

    B) \[\overrightarrow{E}\] on the LHS of the above equation will have a contribution from all charges while q on the RH5 will have a contribution from \[{{q}_{2}}\]and \[{{q}_{4}}\] only

    C) \[\overrightarrow{E}\] on the LHS of the above equation will have a contribution from all charges while q on the RH5 will have a contribution from \[{{q}_{1}},\,{{q}_{3}}\]and \[{{q}_{5}}\] only

    D) Both \[\overrightarrow{E}\]on the LH5 and q on the RHS will have contribution from \[{{q}_{2}}\]and \[{{q}_{4}}\]only

    Correct Answer: B

    Solution :

    (b) \[\overrightarrow{E}\]on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from \[{{q}_{2}}\]and \[{{q}_{4}}\] only. When the point is on the diameter and away from the centre of hemisphere which is charged uniformly and positively, the component of electric field intensity parallel to the diameter cancel out. So, the electric field is perpendicular to the diameter.


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