# Answer the following Electrostatics question

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in JEE A rigid insulated wire frame in the form of a right angled triangle ABC, is set in a vertical plane as shown. Two bead of equal masses m each and carrying charges q1 & q2 are connected by a cord of length l & slide without friction on the wires. Considering the case when the beads are stationary, determine.
(a) The angle α.
(b) The tension in the cord &
(c) The normal reaction on the beads.

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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=60o, we have
Mgsin60o=(T−F)sin60o

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

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