Capacitor partially dipped in water

  • Thread starter Thread starter rohith291991
  • Start date Start date
  • Tags Tags
    Capacitor Water
AI Thread Summary
The discussion revolves around a parallel plate capacitor partially submerged in a liquid with relative permittivity k, focusing on the height change of the liquid when the capacitor is charged and then disconnected from the power source. The initial calculations for the height of the liquid are presented, but confusion arises regarding the forces acting on the water and why the energy changes when the capacitor is disconnected. Participants question the clarity of the problem, particularly the definition of the height ratio before and after charging. The main concern is understanding the mechanism behind the height change and the energy loss in the system. The conversation highlights the need for clearer problem definitions in physics to avoid ambiguity in solutions.
rohith291991
Messages
7
Reaction score
0

Homework Statement


A parallel plate capacitor is dipped in a liquid of relative permittivity k horizontally, so that the upper plate remains outside. A source of emf V volt is used to charge the capacitor through a key K. The key K is first closed and then opened. The ratio of respective heights of the level of the liquid from the lower plate is...

Homework Equations


Adgh=F...for equilibrium...

The Attempt at a Solution


i got the first height as h=(((K^2)-1)q^2)/(2(A^2)(K^2)epsilon0dg)

where q is the charge on the capacitor, A is the area of each plate, d is the density...

now my doubt is how to find the force acting on the water in the second case... won't it be the same as the first case?...why should the height change at all?

i saw the solution and it goes like this.. First the potential energy as a function of h was found and the negative of its derivative wrt h gave the force...but why does the energy vary??/ in fact it was given that the energy in the second case is less than in the first case...y is this so?? how is the energy lost??
 
Last edited:
Physics news on Phys.org
please someone help me out...
 
rohith291991 said:
A parallel plate capacitor is dipped in a liquid of relative permittivity k horizontally, so that the upper plate remains outside. A source of emf V volt is used to charge the capacitor through a key K. The key K is first closed and then opened. The ratio of respective heights of the level of the liquid from the lower plate is...


I think no one has dealt with this problem because it is unclear what the question is. What ratio is being taken here? Is it the ratio of the heights of the liquid between the plates before and after the capacitor has been charged? Is the liquid confined in any way? It reads as if either you omitted some details or the problem isn't well-defined.
 
no the capacitor is charged and the key remains closed...let the height of the water be h1...then the key is opened(the capacitor is no longer connected in the circuit)...let the height now be h2... the ratio h2/h1 is asked... i guess this is quitw clear...my doubt is y wud the heights change at all??
 
also i don't know what u mean by confined...there is a liquid surface that's all...it doesn't matter whether its the surface of a sea or water in a cup...i don't understand y it shud matter...the question seems to be clear enough...
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Charge on inner/outer surfaces of two large parallel conducting plates
Replies
11
Views
3K
Changes in capacitor after dielectric inserted
Replies
17
Views
3K
Material between capacitor plates - viewed as two capacitors in series
Replies
3
Views
1K
Final potential difference of a 2 capacitor system
Replies
3
Views
3K
Uncharged capacitors connected in series
Replies
18
Views
3K
What Is the Direction of Force on a Dielectric Slab in a Capacitor?
Replies
10
Views
4K
Connecting two charged capacitors
Replies
3
Views
3K
Capacitor+work energy theorem problem
Replies
18
Views
2K
How Does Inserting Different Dielectric Materials Affect Capacitor Performance?
Replies
9
Views
2K
Charged capacitors connected in series
Replies
10
Views
3K

Hot Threads

Recent Insights

Back
Top