Solve please!!!! I need help!!
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Hola !!!
this is based on concept of electrostatic ===== × =======× ==============×
if we split the component of Tension, we see in equilibrium condition , Tcos15° = mg { virtical equilibrium
Tsin15° = Fs = KQ²/r² = Q²/4π€°r² { where r = 2lsin∅ , l is the length of string }
then, tan15° = Q²/4π€°r²mg ------(1) initially
now, when , system submerged in liquid .
use Archenemies concept ,
Fnet{ downward in one string } = mg( 1-P/d)
where P is the density of liquid and d is the density of material
hence, similarly ,
tan15° = Fs'/Fnet
tan15° = Q²/4π€mg(1-P/d) -------(2)
from eqn (1) and (2)
1/€° = 1/€(1 -P/d)
€° = €(1 - P/d )
€° = €° ×€' ( 1- P/d ) { here €' is actual dielectric constant of liquid }
€' = 1/( 1- P/d )
= 1/( 1- 0.8/1.6) = 2
hence , dielectric constant =2
Hope it helps you ...............
tysm...................#kundan
this is based on concept of electrostatic ===== × =======× ==============×
if we split the component of Tension, we see in equilibrium condition , Tcos15° = mg { virtical equilibrium
Tsin15° = Fs = KQ²/r² = Q²/4π€°r² { where r = 2lsin∅ , l is the length of string }
then, tan15° = Q²/4π€°r²mg ------(1) initially
now, when , system submerged in liquid .
use Archenemies concept ,
Fnet{ downward in one string } = mg( 1-P/d)
where P is the density of liquid and d is the density of material
hence, similarly ,
tan15° = Fs'/Fnet
tan15° = Q²/4π€mg(1-P/d) -------(2)
from eqn (1) and (2)
1/€° = 1/€(1 -P/d)
€° = €(1 - P/d )
€° = €° ×€' ( 1- P/d ) { here €' is actual dielectric constant of liquid }
€' = 1/( 1- P/d )
= 1/( 1- 0.8/1.6) = 2
hence , dielectric constant =2
Hope it helps you ...............
tysm...................#kundan
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