**Solution: (a) To find angle α**

Let us consider forces acting on bead P as shown in fig. These forces are :

i) Weight mg vertically downwards

ii) Tension Tin the string

iii) Electric force between P and 0 given by

F=(q1)⋅(q2)/4πε0xl2

iv) Normal reaction N_1

The net force along the string is (T−F). Bead P will be in equilibrium, if the net force acting on it is zero. Resolving forces mg and (T−F) parallel and perpendicular to plane AB, we get, when the bead P is in equilibrium,

mgcos60°=(T−F)cosα......(1)

and

N1=mgcos30°+ (T−F)sinα...... (2)

For the bead at Q, we have

mgsin60°=(T−F)sinα...... (3)

and N_2 = mg cos 60° + (T - F) cos C alpha ...... (4)

Dividing Eq. (3) by Eq. (1) , we get

**tanα=tan60°or α=60°**

**(b) Tension in the chord**

Take α=60^{o}, we have

Mgsin60^{o}=(T−F)sin60^{o}

or **T=F+Mg=q1q2/4πε0xl2+Mg**........(5)

**(c) The normal reaction on the beads**

From Eq. (2) we have (since T−F=mg)

**N1= mgcos30°+mgsin60°=2mgcos30°=√3mg**