Five point charges, each of value +q are placed on five vertices of a regular hexagon of side Lm. 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 be zero (as the forces due to individual charges will balances 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.
`vec(F) + vec(f) = 0` i.e. `vec(F) = -vec(f)`
or, `|F| = |f| = (1)/(4pi epsilon_(0))(q xx q)/(L^(2)) = (1)/(4pi epsilon_(0)) (q^(2))/(L^(2))`

`vec(F)_("net") vec(F)_(CO) = (1)/(4pi epsilon_(0))(q^(2))/(L^(2))` along CO
Method : II

In the diagram we can see that force due to charge A and D are opposite to each other
`vec(F)_(DO) + vec(F)_(AO) = 0` ...(i)
Similarly `vec(F)_(BO) + vec(F)_(EO) = 0` ...(ii)
So `vec(F)_(AO) + vec(F)_(BO) + vec(F)_(CO) + vec(F)_(DO) + vec(F)_(EO) = vec(F)_("Net")`
Using (i) and (ii) `vec(F)_("Net") = vec(F)_(CO) = (1)/(4pi epsilon_(0))(q^(2))/(L^(2))` along CO.
(A) The total charge of the rod cannot be considered to be placed at the centre of the rod as we do in mechanics for mass in many problems.
(B) If `a gt gt 1` then `F = (KQ q)/(a^(2))` behaviour of the rod is just like a point charge.