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Question

A particle A having a charge of 5.0×107C 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 above the particle A. Find the angle of thread with the vertical when it stays in equilibrium.
1157800_52fa314c367e472ba897e96e66aa52eb.PNG

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Solution

Let the thread make angle θ with the vertical.

We see that ΔAOB is an isosceles triangle.

Since its vertex angle =θ, base angles are =(90θ2). . Sum of all three angles of a triangle =180)

If we drop a perpendicular from O on AB, it divides the line AB in two equal parts.

Using trigonometry

AB=0.6sin(θ2)m

The magnitude of Coulomb's Force is found from the given expression

F=ke|qAqB|(AB)2

where ke is Coulomb's constant =8.99×109Nm2C2

Inserting various values we get

F=8.99×109×5.0×107×5.0×107(0.6sin(θ2))2

F=6.24×103sin2(θ2)

Using Lami's theorem which relates the magnitudes of three coplanar, concurrent

and non-collinear forces, T,F and W keeping the charged particle B in static

equilibrium and angles between these we get

Fsin(180θ)=mgsin(90+θ2)=Tsin(90+θ2)

Using first equality and simplifying we get

Fsinθ=mgcos(θ2)

Taking g=9.81ms2, inserting various values and rewriting sin θ in terms of half angle we get

6.24×103sin2(θ2)2sin(θ2)cos(θ2)=0.1×9.81cos(θ2)

6.24×1032sin3(θ2)=0.1×9.81

sin3(θ2)=6.24×1032×0.1×9.81

sin3(θ2)=0.00318

(θ2)=sin10.14706

θ=16.9.


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