Question

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is α. The spheres are now immersed in a dielectric liquid of density 800 kg m^−3 and dielectric constant 21. If the angle between the strings remains the same after the immersion, then
(A) electric force between the spheres remains unchanged
(B) electric force between the spheres reduces
(C) mass density of the spheres is 840 kg m^−3
(D) the tension in the strings holding the spheres remains unchanged

Answer 1

It has given that, at equilibrium the angle between the string is α. The sphere are now immersed in a dielectric liquid of density 800 kg/m³ and dielectric constant is 21. after immersion, angle between strings remains the same.

solution : Here net electric force on any sphere is lesser due to force of buoyancy acts on each sphere but force between the sphere remains unchanged because it depends on seperation between sphere and charges.

see the diagram,

let T is the tension on string and m is the mass of each sphere.

at equilibrium,

vertical component of tension = mg

Tcosα/2 = mg .....(1)

and horizontal component of tension = F

Tsinα/2 = F......(2)

so dividing equation (2) by (1) we get,

⇒tanα/2 = mg/F .......(3)

after immersed in dielectric liquid. as there is no change in angle α.

so, Tcosα/2 = mg - Vρg ........(4)

and Tsinα/2 = F/K ......(5)

now after dividing equation (5) by (4) we get,

tanα/2 = (mg - Vρg)/F/K

from equation (3)

mg/F = (mg - Vρg)K/F

⇒mg/K = mg(1 - ρ/d)/1 [ as Vρg = mg(ρ/d) ]

⇒1/K = 1 - ρ/d

here K = 21 , ρ = 800

so, 1/21 = 1 - 800/d

⇒d = 840

Therefore density of spheres is 840 kg/m³

Therefore the correct options are (A) and (C)

Answer 2

$\bf Answer :- \: \: \: Option \: (A,C)$

$\bf T \: cos \: θ = mg \: (i)$

$\bf T \: sin \: θ = qε \: (ii)$

$\bf qε = mg \: t an \: θ \: (iii)$

When medium is introduced

$\bf T ^{1} \: cos \: θ=mg - f _ 4</p><p>$

$\bf T ^{1} \: cos \: θ=mg \: \bigg(1 - \frac{ ρ _1}{ρ _s} \bigg ) \: (iv)$

$\bf T ^{1} \: sin \: θ= \frac{qε}{K} \: \: (v)$

Divide equation (v) and (ii)

$\bf \frac{T^{1} }{T } = \frac{q∑}</p><p>{kq∑}$

$\bf T ^{1} = \frac {T}{k} = \frac { T}{21} \: (vi)$

Divide equation (iv) and (v)

$\bf \frac{q\epsilon}{k} = mg \bigg (1 \: - \: \frac{\rho _ 1}{ \rho_s} \bigg ) tan \: θ$

$\bf qε=kmg \ \bigg (1 \: - \: \frac{\rho _ 1}{ \rho_s} \bigg ) tan \: θ  \: \: (vii)$

equate eqn (vii) and (iii)

$\bf ρ_s = 840 \: kg/ {m}^{3}$