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,
(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