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.You need to login to perform this action.
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