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?

Explanation

Method-I : If there had been a sixth charge +q at the remaining vertex of hexagon, force due to all the six charges on -q at O would have been zero (as the forces due to individual charges will balance each other), i.e., `vec(F_(R))=0`
Now if `vec(f)` is the force due to sixth charge and `vec(F)` due to remaining five charges.
From `vec(F)+vec(f)=0` i.e. `vec(F)=-vec(f)`
or, `|F|=|f|=1/(4pi epsi_(0)) (qxxq)/L^(2)=1/(4pi epsi_(0)) q^(2)/L^(2)`
`vec(F)_("Net")=vec(F)_(OD)=1/(4pi epsi_(0)) q^(2)/L^(2)` along `OD`

Method-II : In the diagram, we can see that force due to charge A and D are opposite to each other
`vec(F)_(OF)+vec(F)_(OC)=vec(0)` ....(i)
Similarly `vec(F)_(OB)+vec(F)_(OE)=vec(0)` ...(ii)
So `vec(F)_(OF)+vec(F)_(OB)+vec(F)_(OC)+vec(F)_(OD)+vec(F)_(OE)=vec(F)_("Net")`
Using (i) and (ii) `vec(F)_("Net")=vec(F)_(OD)=1/(4pi epsi_(0)) q^(2)/L^(2)` along OD.