Question

Capacitors question , please help :p

Answer 1

Interesting Question :D 
Q : A parallel- plate  is located horizontally so that one of its plates is submerged into liquid while the other is over its surface .The permitivity of the liquid is equal to ∈ , its density is р . To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface density σ ? 

Solution :
  Final Answer :  [σ^(2)*(k^(2)-1)]/ [2ε(o)pgk^{2}]
Steps : 
1) 
Q: Why Level of liquid will rise ?
=> Due to Polarization of charges or we can say there will be induced charges on the Dielectric (liquid) .These induced charges will be attracted by surface charges of parallel plate capacitor .

2)
Now,
Let the rise in level of liquid be 'h'. 
There will be induced charges on both sides of dielectric whose sign is opposite  to the  plate charge facing .
                   
Induced charge ,σ(i) = σ(1-1/k)   -----(1) 

where σ is surface charge density of parallel plate capacitor .
k -> Dielectric constant (Relative Permitivity) = ε/ε(o) 
where ε -> Absolute permitivity  of liquid .

3) There will be an upward force which pulls the mass of liquid upward .
Let the mass of liquid pulled upward be 'm' .
m= pAh ,---- (2) 
where A is surface area of liquid pulled upward . 

We know, when liquid reaches height h ,it comes in equilibrium .
E(plates) = A*(σ/2ε(o)) *2 
E(induced)  = A *(σ(i)/2ε(o))
where E -Electric  Field 

F(up) = mg 
=> qE(net)  =  pAhg 
=> q   *(E(plates) + E(induced)) =pAhg
=> σ(i)A  * ( A*(σ/2ε(o)) *2 - σ(i) A/(2ε(o)) =pAhg 
On solving and substituting value of σ(i)  we get ,  

h = σ^(2) (k^(2)-1) /(2ε(o) pgk^(2) ) 
Seems,like you  provide Absolute permitivity ε,so just substititute 
k=ε/ε(o) 

This is Really Famous Question . 

But , you will get wrong answer if you apply conservation of energy as there will be a dissipation of energy when liquid rises up .
IF your book give different answer other than this ,then that  answer is wrong  cause that is obtained by Conservation of Energy Method .