Question

Five identical charges are kept at live vertices of a regular hexagon, Match the following two columns at centre of the hexagon, If in the given situation electric field at centre is E, Then

Answer 1

(A) In given situation field at center is E towards A which due to charge $q_D$
Now if charge B is removed then then field at center will be E towards A and E towards B
Now resulting field will be $2E\cos \theta /2=2Ecos30=\sqrt{3}E$
A->4
(B)If charge at C is removed then field at center will be E towards A and E towards C
resultant field $2E\cos 2\theta /2=E$
B->2
(C)filed at center initially was due to charge at D now if it is removed then field will become zero.
C->3
(D)Since when charge at C is removed field at center is E towards B
and field at center due charge at B is balancing the field due to charge at E therefore, when charge at B is also removed then field due E will also have influence and which is towards B therefore, both the fields will add up and resultant field at center will be 2E
D->1