Five point charge each of value +q are placed on five vertices of a regular hexagon of side a metre. What is the magnitude of the force on a point charge of value `-q` coulomb placed at the centre of the hexagon?

Options

A. `(1)/(pi epsilon_(0))((q)/(L))^(2)`
B. `(2)/(pi epsilon_(0))((q)/(L))^(2)`
C. `(1)/(2 pi epsilon_(0))((q)/(L))^(2)`
D. `(1)/(4 pi epsilon_(0))((q)/(L))^(2)`

Answer

D

Explanation

If there had been a sixth charge +q at the remaining vertex of hexagon, force due to all the six charge on -q at O will be zero.

Now if f is the force due to the sixth charge and F due to the remaining five charge, then
`vec(F)+vec(f)=0, i.e. vec(F)=-vec(f)`
`|vec(F)|=|vec(f)|= (1)/(4 pi epsilon_(0)) (qxxq)/(L^2)= (1)/(4 pi epsilon_(0))((q)/(L))^(2)`.