Question

What is the resultant force acting on the charged bead?

Answer 1

The positive charge at A will pull the negaitvely charged bead at C towards A.

Similarly the one at B will pull it towards B.

These two forces $F_1$ and $F_2$ respectively, will have their components along CO and perpendicular to CO along the opposite directions.

Magnitude wise both $F_1,F_2$ are equal

The perpendicular components will get cancelled and the components along CO will get added up.

The cosine of angle ACO is $\frac{OC}{AC}$

The cosine of angle BCO is $\frac{OC}{BC}$

The resultant force = $\frac{F_1\times OC}{AC}+\frac{F_2\times OC}{BC}$---------------------------------------------------------(1)

Since it is given that $x_0<<R$, OC is negligible and the value of resultant foce becomes almost zero.

Answer:- Negligible unbalanced force is acting on the bead along CO