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

Answer:

ANSWER

(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

Given:

Four identical point charges are placed on the vertices of a regular hexagon of side a such that the net electric field at the centre is zero.

Now , one of the charge is removed.

To find:

Electrostatic field intensity at the centre.

Calculation:

We can easily understand that the four identical charges had been placed at diametrically opposite vertices such that the field intensity nullified each other and the net intensity was zero at the centre.

But, due to removal of one of the charge there will be a net field intensity at the centre.

The net field intensity at the centre will be equal to the field intensity which would have been provided by the removed charge.

Net intensity:

$\therefore \: E_{net} = \dfrac{kq}{ {d}^{2} }$

$= > \: E_{net} = \dfrac{kq}{ {a}^{2} }$

Putting value of Coulomb's Constant:

$= > \: E_{net} = \dfrac{q}{4\pi \epsilon_{0} {a}^{2} }$

So, final answer is :

$\boxed{ \red{ \bold{E_{net} = \dfrac{q}{4\pi \epsilon_{0} {a}^{2} } }}}$