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---?

(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,

wikipedia

3 Answers
Aug 9, 2017

Let #O# be point of suspension of charged particle #B#. Charged particle #A# is fixed and particle #B# is held in equilibrium under forces as shown in the fig

  1. Coulomb's Force of repulsion #F# along #AB#
  2. Tension #T# in the silk thread.
  3. Weight acting downwards #mg#
    http://vidyadrishti.com

Let the thread make angle #theta# with the vertical.
We see that #Delta AOB# is an isosceles triangle.
Since its vertex angle #=theta#, base angles are #=(90^@-theta/2)#. (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=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, #T,F and W# keeping the charged particle #B# in static equilibrium and angles between these we get
#F/(sin(180-theta))=(mg)/(sin(90+theta/2))=T/(sin(90+theta/2))#

Using first equality and simplifying we get

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

Taking #g=9.81ms^-2#, inserting various values and rewriting #sin theta# in terms of half angle we get
#((6.24xx10^-3)/ (sin^2(theta/2)))/(2sin(theta/2)cos(theta/2))=(0.1xx9.81)/(cos(theta/2))#
#=>(6.24xx10^-3)/(2sin^3(theta/2))=0.1xx9.81#
#=>sin^3(theta/2)=(6.24xx10^-3)/(2xx0.1xx9.81)#
#=>sin^3(theta/2)=0.00318#
#=>sin(theta/2)=0.14706#
#=>(theta/2)=sin^-1 0.14706#
#=>theta=16.9^@#

Aug 9, 2017

Answer:

#theta~~16.9164^@~~0.29525# rad

Explanation:

enter image source here

It is clear from the figure that #Delta OAB# is an isosceles triangle, so

#2beta+theta=180^@#
#=>beta=90^@-theta/2#

Also,

#alpha+beta=90^@#
#=>alpha=90^@-(90^@-theta/2) = theta/2 ........................(1)#

From figure it is clear that in equilibrium position,

#mg*sin(theta)=F_e*cos(alpha)#

#=>mg*sin(2alpha)=F_e*cos(alpha)#

#=>2mg*sinalpha*cancelcolor(red)(cosalpha)=F_e*cancelcolor(red)(cos(alpha))#

#=>2mg*sinalpha=k_e(Q_AQ_B)/r^2=k_e(Q^2)/r^2...................(2)#

Now from #DeltaOAB#,

#costheta=(l^2+l^2-r^2)/(2*l^2)=1-(r^2)/(2l^2)#

#=>r^2=2l^2(1-costheta)=2l^2(1-cos(2alpha))#

Put the value of #r^2# in equation #(2)#,

#=>2mg*sinalpha=k_e(Q^2)/(2l^2(1-cos(2alpha)))=k_e(Q^2)/(2l^2(2-2cos^2alpha))#

#=>2mg*sinalpha=(k_eQ^2)/(4l^2sin^2alpha)#

#=>sinalpha=root(3)((k_eQ^2)/(8mgl^2))=root(3)((8.9875517873681764*10^9*(5*10^-7)^2)/(8*0.1*9.80665*(0.3)^2))= sin(theta/2)#

#=>theta=2*sin^-1(root(3)((8.9875517873681764*10^9*(5*10^-7)^2)/(8*0.1*9.80665*(0.3)^2)))#

#=>theta~~16.9164^@~~0.29525# rad

Aug 9, 2017

Answer:

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

Explanation:

Let #O# be point of suspension of charged particle #B#. Charged particle #A# is fixed and particle #B# is held in equilibrium under forces as shown in the fig

  1. Coulomb's Force of repulsion #F# along #AB#
  2. Tension #T# in the silk thread.
  3. Weight acting downwards #mg#
    http://vidyadrishti.com

As three forces are in equilibrium, sum of vertical components of #T and F# are equal and opposite to #mg# and Horizontal components of #T and F# are equal and opposite. From the figure we see that #F# makes an angle# =(90-theta/2)# with the vertical and #T# makes an angle #=theta# with the vertical.

We get
#Tcostheta+Fcos(90-theta/2)=mg#
#=>Tcostheta+Fsin(theta/2)=mg# ..... (1)

#Tsintheta=Fsin(90-theta/2)# .....(2)
From (2)
#=>T=1/sin theta Fcos(theta/2)#
Inserting this value of #T# in (1) we get

#(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