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If a charge q is placed at each vertex of a regular polygon, the net electric field at its centre is zero. Show it. |
Answer» Solution : The distance of the centre of a regualr polygon from each vertex will be same. Therefore, `|vec(E_(1))|=|vec(E_(2))|=|vec(E_(3))|` in (a) `|vec(E_(1))|=|vec(E_(2))|=|vec(E_(3))|=|vec(E_(4))|` in (b) and `|vec(E_(1))|=|vec(E_(2))|=|vec(E_(3))|=|vec(E_(4))|=|vec(E_(5))|` in ( c ) The angle between any two consecutive field vectors for each of a polygon is also same. Hence, `vec(E_(1))+vec(E_(2))+vec(E_(3))+......+vec(E_(n))=vec(0)` where n = 3, 4, 5, 6.............. Note : `vec(E)_(1)+vec(E)_(2)+............+vec(E)_(n)iff vec(E)_(1)+vec(E)_(2)+............+vec(E)_(n-1)=-vec(E)_(n)` Hence, if a charge q is PLACED at each vertex EXCEPT one vertex of regular polygon, then the net electric field at its centre, distant r from each vertex is `(1)/(4pi epsilon_(0))(q)/(r^(2))`, directed TOWARDS or away from the empty vertex depending on whether q is positive or negative. |
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