- #1
RingNebula57
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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?
##C = \frac{\epsilon_0 \cdot S}{d}##
##C'=\frac{\epsilon_0 \cdot S}{\eta^2 d} ##
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...
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...