LC circuit -- Calculate the work done

  • #1
RingNebula57
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2
1. The problem statement, all variables and given/known data
In an oscillating circuit consisting of of a parallel-plate capacitor and an inductance coil with negligible active resistance the oscillations with energy ##W## are sustained. The capacitor plates were slowly drawn apart to increase the scillation frequency ##\eta##-fold. What work was done in the process?

Homework Equations


##C = \frac{\epsilon_0 \cdot S}{d}##
##C'=\frac{\epsilon_0 \cdot S}{\eta^2 d} ##

The Attempt at a Solution


So ohm's law is written like this: (before the stretching of the capacitor)

##L \cdot \ddot{q} = -\frac{q}{C}##
This implies:
##\omega^2=\frac{1}{LC}##
##q_{(t)}=q_m \cdot sin(\omega t)##

The electrostatic force exerted by one plate on the other is:

## F = E_{(t)} \cdot q_{(t)} ##
## E_{(t)} = \frac{\sigma}{2\epsilon_0} = \frac{q_{(t)}}{2\epsilon_0S} ## ##\Bigg\} ## ##F = \frac{q_{(t)}^2}{2\epsilon_0S}=\frac{q_m^2 \cdot sin(\omega t)^2}{2\epsilon_0S}##

Because the streching of the capacitor happened slowly it means that we can consider the average force over one period:

##\bar{F}=\frac{q_m^2}{4\epsilon_0S}##
So , the work done is then:
## \Delta W= \bar{F}\cdot d\cdot (\eta^2-1) = \frac{q_m^2d}{4\epsilon_0S} \cdot (\eta^2-1)##

Now,
##W=\frac{q_m^2}{2C}=\frac{q_m^2d}{2\epsilon_0S}##

From here results that :
##\Delta W= \frac{W}{2} \cdot (\eta^2-1)##

This formula implies that the initial charge ##q_m## is not conserved. Is it ok? It looked odd to me...
 
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  • #2
RingNebula57 said:
##\bar{F}=\frac{q_m^2}{4\epsilon_0S}##
I believe this is correct.

So , the work done is then:
## \Delta W= \bar{F}\cdot d\cdot (\eta^2-1) = \frac{q_m^2d}{4\epsilon_0S} \cdot (\eta^2-1)##
This assumes that ##\bar{F}## remains constant while the plates are separated. But as you found, ##q_m## is not conserved. So, the average force is not constant.

Now,
##W=\frac{q_m^2}{2C}=\frac{q_m^2d}{2\epsilon_0S}##
Yes

For notational convenience, I suggest letting ##x## stand for the distance between the plates. Consider the work done by the average applied force for an infinitesimal increase in separation, ##dx##.
 
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  • #3
TSny said:
I believe this is correct.

This assumes that ##\bar{F}## remains constant while the plates are separated. But as you found, ##q_m## is not conserved. So, the average force is not constant.

Yes

For notational convenience, I suggest letting ##x## stand for the distance between the plates. Consider the work done by the average applied force for an infinitesimal increase in separation, ##dx##.
Thank you!

I solved it. Obtaining that :

##\Delta W = W \cdot (\eta-1)##
 
  • #4
RingNebula57 said:
I solved it. Obtaining that :

##\Delta W = W \cdot (\eta-1)##
Looks good!
 
  • #5
RingNebula57 said:
Thank you!

I solved it. Obtaining that :

##\Delta W = W \cdot (\eta-1)##
Please help me with the solution...as I am g
etting a logarithmic term while integration of 1/x
 
  • #6
Anubhav said:
Please help me with the solution...as I am g
etting a logarithmic term while integration of 1/x
Please start a new thread with your question. This thread here is 5 years old. Thank you.
 
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