Q: Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting , mass less string of same length . At equilibrium the angle between the strings is θ . 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

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Sol: In air ,

T cosθ = m g …(i)

T sinθ = q E …(ii)

on dividing , qE = m g tanθ

In liquid ,

T’cosθ = m g – F_{B}

$T’ cos\theta = m g(1+\frac{\rho_l}{\rho_s})$ …(iii)

$T’ sin\theta = \frac{qE}{K}$ …(iv)

So , $\frac{T’}{T} = \frac{1}{K}$

$T’ = \frac{T}{K} = \frac{T}{21}$

On dividing (iv) by (iii)

$\frac{qE}{K} = m g (1-\frac{\rho_l}{\rho_s})tan\theta $

$qE = K m g (1-\frac{\rho_l}{\rho_s})tan\theta$

Solving , ρ_{s} = 840 kg/m^3