37 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.approach 1 :
force on dielectric * 2a = 1/2mv^2
gives root2 times of given ans
I'm sure it's a tough question but could we get some more pixels?
seems clear as day to me
Ok better
approach 2 : 1/2 mv^2 + Ui + Wb = Uf + H + Work done by force on dielectric
idk how to figure out heat or this even works
Capacitance changes, charge is pushed, so ig there would be some heat. (Although, my physics is weak but they are my 2 cents ;_;)
There used to be a formula for force on a dielectric I think.
Would that be helpful?
Oh right. The force on dielectric when connected to constant voltage source is constant, hence you just have to use 2ax=v^2-u^2; where v=0, u=initial velocity of dielectric=initial velocity of bullet and a is acceleration due to force by cap. and x is length of dielectric.
x = 2a right ?
2ax = -u^2
substitute x=2a ( because dielectric is pushed out ) and value of accn we get some expression for v
problem 1 : it has (1-k) term which is -ve thus making velocity imaginary
problem 2 : if we ignore the 1-k and make it k-1 , still the answer is root2 times of given ans
Nope. x=a (Force is applied when dielectric starts to go out of capacitor).
Also the - term cancels out. Imma send the solution in a bit
If we conserve energy, there will be two expression for kinetic, initally for bullet and then fir dielectric, right?
@SirLancelotDuLac
Heat is liberated when charge is pushed tho (due to changing capacitance)
I dont think we can calc heat produced
Energy conservatn isnt useful here
This works, I think.
Can you tell how we calxulate the force here?
Its better to remember ig, but to calculate it:
1. Consider the situation when x portion of the dielectric is inside the capacitor and find the potential energy
2. Use F=-dU/dx to calculate force.
The results are different for battery being connected (which is constant) and battery being disconnected.
Oh yea yea i remember, there is an example in HC VERMA given calculating this force
d is the separation between plates right?
Yep.
But in the question, d isnt given
That's true, but the answer key seems to use it, ig they forgot to mention. :/
Oh
Can you explain again why we took x=a and not 2a
The force only begins to act when the dielectric starts to leave the capacitor. Between that and when it completely leaves (i.e., when the force starts and stops acting), it moves a distance of a.
Ohh i get it
Thanks bro
huh?
force is constant no matter the length of dielectric inside capacitor
Yep. That's why 2ax=v^2-u^2 is used.
As for the x=a thing.
so u mean to say that when entire dielectric is inside capacitor no force is acting on it?
oh
fringing force acts on the length of dielectric outside the capacitor. so when dielectric is completely inside no force acts on it.
@SirLancelotDuLac confirm
Yep.
Confirmed
Alrr
@SirLancelotDuLac rn
Fringing force mean?
At the end of capacitor the field lines tend to bend . The force applied due to these bending / fringing field lines is termed as fringing force
Btw what u doing now? Drop or clg
Oh okay
+solved @SirLancelotDuLac
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