Two identical charged spheres of material density `rho`, suspended from the same point by inextensible strings of equal length make an angle `theta` between the string. When suspended in a liquid of density `sigma` the angle `theta` remains the same. The dielectric constant K of the liquid is

Options

A. `(rho)/(rho-sigma)`
B. `(rho-sigma)/(rho)`
C. `(rho)/(rho+sigma)`
D. `(rho+sigma)/(rho)`

Answer

A

Explanation

Initially,the forces acting on each sphere are as follows:
(i)Weight,W=mg
(ii)Tension in the string,T
(iii)Electrostatic force of repulsion ,F
For equilibrium of sphere
`Tsin((theta)/(2))=F`.....(i)
and `Tcos((theta)/(2))`=mg......(ii)
Divide (i) by (ii),we get
`tan((theta)/(2))=(F)/(mg)`......(iii)
When they are suspended in a liquid of density `rho` and dielectric constant K,the angle `theta` and hence the distance between the spheres remains the same,Therefore,
Force of repulsion will become
`F.=(F)/(K)`.......(iv)
and weight will become
mg.=`mg-Vrhog`
`=mg(1-(rho)/(sigma)/(rho))`.........(v)
Let T. is the new tension in string. For equilibrium of sphere,
`T.sin((theta)/(2))=F`.........(vi)
`T.cos((theta)/(2)=`mg..............(vii)
Divide (vi) by (vii),we get
`tan((theta)/(2))=(F.)/(mg.)=(F//K)/(mg(1-sigma)/(rho))`.......(viii)
(Using (iv) and (v))
From equations (iii) and (viii),we get
`K=(1)/(1)/(1-(sigma)/(rho))=(rho)/(rho-sigma)`