Question

A particle A, having a charge of 5.0*10^-7C is fixed in a vertical wall. A second particle B of mass 100 g and having equal charge is suspended by a silk thread of length 30 cm from the wall. The point of suspension is 30 cm above the particle A---?

Answer 1

your question is incomplete. A complete question is ---> A particle, A, having a charge of 5.0*10^-7C is fixed in a vertical wall.A second particle B of m=100 g and having equal charge is suspended by a silk thread of length 30 cm from the wall.The point of suspension is 30 cm above the particle.Find the angle of the thread when it stays in equilibrium.

solution : Let angle between thread and the wall in equilibrium condition is ∅

here, OA = OB = 30cm = 0.3m

angle OAB = angle OBA = 90° -∅/2

from figure,

OA/sin(90° - ∅/2) = AB/sin∅

or, OA/cos∅/2 = AB/(2sin∅/2 cos∅/2)

or, 0.3 = AB/(2sin∅/2)

or, AB = 0.6sin(∅/2)

Now from Coulomb's law, F = Kq²/AB²

= 9 × 10^9 × (5 × 10^-7)²/(0.6sin∅/2)²

= 9 × 10^9 × 25 × 10^-14/0.36sin²(∅/2)

=6.25 × 10^-3/sin²(∅/2) .....(1)

using Lami's theorem,

F/sin(180° -∅) = mg/sin(90°+∅) = T/sin(180° - ∅/2)

or, F/sin∅ = mg/cos(∅/2) = T/sin(∅/2)

using F/sin∅ = mg/cos(∅/2)

6.25 × 10^-3/sin²(∅/2)sin∅ = 0.1 × 10/cos(∅/2)

or, 6.25 × 10^-3/(2sin³∅/2) = 1

or, sin³(∅/2) = 3.125 × 10^-3

or, sin³(∅/2) = 3125/10^6

or, sin(∅/2) = 14.6/10² ≈ sin(17/2)

hence, ∅ = 17°