# 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

#### Answer:

#### Explanation:

It is clear from the figure that

Also,

From figure it is clear that in equilibrium position,

Now from

Put the value of

#### Answer:

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