Four equal charges of value +Q are placed at any four vertices of a regular hexagon of side a. By suitably choosing the vertices, what can be the maximum possible magnitude of electric field at the centre of the hexagon?

(A) `(Q)/(4piin_(0)a^(2))`
(B) `(sqrt2Q)/(4piin_(0)a^(2))`
(C) `(sqrt3Q)/(4piin_(0)a^(2))`
(D) `(2Q)/(4piin_(0)a^(2))`

Options

A. `(Q)/(4piepsilon_(0)a^(2))`
B. `(sqrt(2)Q)/(4piepsilon_(0)a^(2))`
C. `(sqrt(3)Q)/(4piepsilona^(2))`
D. `(2Q)/(4piepsilon_(0)a^(2))`

Answer

c

Explanation

To maximize the elctric field at the centre both charges should be placed at adjecent vertices.
Net field due to charge at A and D is zero here.
Total System of electric field due to the charges is
`thereforeE_("total")=sqrt(E_(F)^(2)+E_(E)^(2)+2E_(F)E_(E)cos60^(@))`
`=sqrt(2E^(2)+2E^(2)xx(1)/(2)) " " (becauseE_(F)=E_(E)=E)`
`=sqrt(2E^(2)+E^(2))=sqrt(3E^(2))orE_("total")=sqrt(3)E`
So , maximum electric field is
`E_("total")=sqrt(3)xx(1)/(4piepsilon_(0))(Q)/(a^(2))` (using(i))