# 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---?

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(continued) Find the angle of the thread with the vertical when it stays in equilibrium.

Forces acting on the particle are

(i) weight mg downward

(ii) tension T along the thread

(iii) electric force of repulsion F

Coulomb's Law,

(continued) Find the angle of the thread with the vertical when it stays in equilibrium.

Forces acting on the particle are

(i) weight mg downward

(ii) tension T along the thread

(iii) electric force of repulsion F

Coulomb's Law,

##### 3 Answers

Let

- Coulomb's Force of repulsion
#F# along#AB# - Tension
#T# in the silk thread. - Weight acting downwards
#mg#

Let the thread make angle

We see that

Since its vertex angle

If we drop a perpendicular from

#AB=2xx[0.3xxsin(theta/2)]#

#AB=0.6sin(theta/2)m#

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

#F = k_e (|q_A q_B|)/ (AB)^2 #

where#k_e# is Coulomb's constant#= 8.99×10^9 N m^2 C^-2#

Inserting various values we get

#F = 8.99×10^9xx (5.0xx10^-7xx5.0xx10^-7)/ (0.6sin(theta/2))^2 #

#=>F = (6.24xx10^-3)/ (sin^2(theta/2)) #

Using Lami's theorem which relates the magnitudes of three coplanar, concurrent and non-collinear forces,

Using first equality and simplifying we get

#F/(sintheta)=(mg)/(cos(theta/2))#

Taking

#### Explanation:

It is clear from the figure that

Also,

From figure it is clear that in equilibrium position,

Now from

Put the value of

Alternate solution steps if you don't want to use Lami's Theorem

#### Explanation:

Let

- Coulomb's Force of repulsion
#F# along#AB# - Tension
#T# in the silk thread. - Weight acting downwards
#mg#

As three forces are in equilibrium, sum of vertical components of

We get

From (2)

Inserting this value of

#(1/sin theta Fcos(theta/2))costheta+Fsin(theta/2)=mg#

#=>F (cos(theta/2)costheta+sin thetasin(theta/2))=mgsin theta#

Using the trigonometric identity we get

#F cos(theta-theta/2)=mgsin theta#

#=>Fcos(theta/2)=mgsin theta#

Proceed as in other solution