3 problems that will clear your concepts on capacitors
1. A cylindrical layer of dielectric with permittivity 
is inserted into a cylindrical capacitor to fill up all the space between the electrodes. The mean radius of the electrodes equals R, the gap between them is equal to d, with d 
R. The constant voltage V is applied across the electrodes of the capacitor. Find the magnitude of the electric force pulling the dielectric into the capacitor.
Solution :
Let the length of cylindrical layer of dielectric inserted in capacitor is x.
Plates are seperated by a distance d & mean radius is R such that 
.
So, capacitance of cylindrical capacitor of mean radius R and distance between two plates d is
approximately equal to 
Here, we have dielectric upto height x.
So, Resultant capacitance = 
Electrostatic energy
Force on dielectric
Negative sign denotes that force is attractive.
2. Inside a parallel-plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to 
of the gap width. When the plate is absent the capacitor capacitance equals 
. First, the capacitor was connected in parallel to a constant voltage source producing V = 200 V, then it was disconnected from it, after which the plate was slowly removed from the gap. Find the work performed during the removal, if the plate is
(a) made of metal (b) made of glass.
Solution :
Before the introduction of dielectrics
Capacitance 
After the introduction of dielectrics
Now capacitor is connected to constant voltage source = V
So, capacitor will be immediately charged to
After this voltage source is disconnected.
Now, capacitor is isolated.
For isolated capacitor charge is constant.
Electrostatic energy stored in capacitor with dielectric
When dielectric is taken out, some work is done on the capacitor. So, this work will be stored in capacitor and hence, the electrostatic energy of capacitor will increase.
We can calculate the work performed during removal of dielectric by just calculating the difference in electrostatic energy after and before the removal of dielectric.
Work performed = 
If dielectric is made of metal then
3. A parallel-plate capacitor is located horizontally so that one of its plates is submerged into liquid while the other is over its surface. The permittivity of the liquid is equal to ![clip_image002[1]](http://iit.fundoobanda.com/wp-content/uploads/2009/08/clip_image00211.gif "clip_image002[1]"){kind=small}
, its density is equal to 
. To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface density 
?
Solution :
When capacitor is given charge density ![clip_image057[1]](http://iit.fundoobanda.com/wp-content/uploads/2009/08/clip_image0571.gif "clip_image057[1]"){kind=small}
. Let us assume than height of liquid raised is h.
Initial height of liquid = 0
So, Electrostatic energy of capacitor = 
After liquid rises to height h
Electrostatic energy
Now, gravitational potential energy acquired by liquid
We must take the height of centre of mass in P.E expression.
Since, capacitor is isolated.
So, its initial electrostatic energy will be equal to sum of electrostatic energy and gravitational potential energy of liquid after it attains height h.
So, 
…………(1)
![clip_image063[1]](http://iit.fundoobanda.com/wp-content/uploads/2009/08/clip_image0631.gif "clip_image063[1]"){kind=small}
Solving the equation
we get
[...] 3 problems that will clear your concepts on capacitors [...]
sir i would like to comment on the third problem
third problem is a problem in I.E Irodov 3.144
and also a solved problem in H.C Verma
Now i would like to know how the problem can be solved through
‘Force’ approach,that would clarify the concepts clearly
Thank you
we can discuss the force approach.water gets polarised in electric field produced by free charges on capacitor plates.The bound charges appears on the outer surface of water.the free charges on upper plates will attract the bound charges on the surface of water.Since it is the work done by electric field on the dielectric ie water.A part of electrostatic energy will get converted into gravitational potential energy. .
Good content, hard to find a quality site like this one anymore. Thanks
hi
in reply 3
why there has only one force act on the bound charges on the surface ? (one force from free charge on upper plate)
why not free charge on the lower plate and bound charge on the buttom surface act it too ?
sorry for my english . i’m thai student