Question #9efd0

1 Answer
Aug 24, 2017

Let each charged sphere have a charge #q#, have a mass #m# and make an angle #theta# with the vertical.

In the initial position of equilibrium electrical force of repulsion #F# and weight of each sphere is connected through expression

#F/(mg)=tantheta#

Using Coulomb's Law and other values we write this as

#(k_eq^2/d^2)/(mg)=(d/2)/(l^2-(d/2)^2)^(1/2)#
#=>q=((mg)/(2k_e) d^3/(l^2-(d/2)^2)^(1/2))^(1/2)#

As the charge starts leaking let #x# be instantaneous distance between the spheres. The above expression reduces to
#q=sqrt((mg)/(2k_e)) x^(3/2)/(l^2-(x/2)^2)^(1/4)#

Differentiating both sides with respect to time #t# we get
#(dq)/dt=sqrt((mg)/(2k_e))d/dt[ x^(3/2)/(l^2-(x/2)^2)^(1/4)]#
#=>#
#(dq)/dt=sqrt((mg)/(2k_e))[ (3/2x^(1/2))/(l^2-(x/2)^2)^(1/4)+((x^(3/2))(-1/4)(-x/2))/(l^2-(x/2)^2)^(-5/4)]dx/dt#

#(dq)/dt# is rate of leakage of charge and velocity #dx/dt=v#. Above expression becomes
#1/v(dq)/dt=sqrt((mg)/(2k_e))[ (3x^(1/2))/(2(l^2-(x/2)^2)^(1/4))+(x^(5/2))/(8(l^2-(x/2)^2)^(-5/4))]#

is the required expression.