Question

Four identical point charges, each of charge q, are placed on the vertices of a regular hexagon of side a such that the net electric field at centre of hexagon is zero. If one of the charge is now removed, then the electric field at centre of hexagon is

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 2Ecosθ/2=2Ecos30=

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 2Ecos2θ/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

Answer 2

Answer:

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 2Ecosθ/2=2Ecos30=

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 2Ecos2θ/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

Explanation:

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