Question

Two equal charges are suspended with non-conducting wires of equal length from the same point. Now, the whole system is taken inside a liquid (water). If the angle between the threads remains same, then calculate the relative permittivity of water. Relative density of both charged particles is $\sigma$.

• A. $\frac{\sigma }{\sigma -1}$
• B. $\frac{\sigma }{1+\sigma }$
• C. $\frac{1-\sigma }{\sigma }$
• D. $\frac{1}{\sigma -1}$

Answer 1

The correct option is A $\frac{\sigma }{\sigma -1}$



From FBD of particle,
$T\cos \theta =mg\left(1\right)$
$T\sin \theta =F_e\left(2\right)$
Divide Eqn. $\left(2\right)÷\left(1\right)$
$\tan \theta =\frac{F_e}{mg}\left(3\right)$

In water,


$T\sin \theta =\frac{F_e}{{\in }_r}--\left(4\right)$
$T\cos \theta =mg-F_b--\left(5\right)$
$\Rightarrow \tan \theta =\frac{F_e}{{\in }_r(mg-F_b)}\left(6\right)$

From (3) and (6)
$\frac{F_e}{mg}=\frac{F_e}{{\in }_r[mg-F_b]}$
$\Rightarrow {\in }_r=\frac{v{\rho }_{p}g}{v{\rho }_{p}g-v{\rho }_{w}g}$
$\Rightarrow {\in }_r=\frac{{\rho }_p}{{\rho }_p-{\rho }_w}[\sigma =\frac{{\rho }_p}{{\rho }_w}]$
$\therefore {\in }_r=\frac{\sigma }{\sigma -1}$