Question

Two identical charged spheres are suspended by strings of equal length. The strings make an angle of 30${}^{\circ }$ with each other. When suspended in a liquid of density $0.8g/cc$, the angle remains the same. The dielectric constant of the liquid is :

[density of the material of sphere is $1.6g/cc$]

• A. $2$
• B. $4$
• C. $2.5$
• D. $3.5$

Answer 1

Answer: (A)

For equilibrium of any sphere, $T\cos {15}^o=mg$ and $T\sin {15}^o=F_e$ where $F_e=$ electrostatic force between the charged spheres.

$\therefore \tan {15}^o=\frac{F_e}{mg}....\left(1\right)$

When each sphere is immersed in liquid , the angle still ${30}^o$. But the electrostatic force becomes $\frac{F_e}{K}$ where K is the dielectric constant of the liquid and the effective weight becomes $mg(1-\frac{0.8}{1.6})=mg/2$

now, $\tan {15}^o=\frac{F_e/K}{mg/2}....\left(2\right)$

$\left(1\right)/\left(2\right)\Rightarrow 1=2/K$

$\therefore K=2$