# LC circuit

1. Oct 31, 2016

### RingNebula57

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?

2. Relevant equations
$C = \frac{\epsilon_0 \cdot S}{d}$
$C'=\frac{\epsilon_0 \cdot S}{\eta^2 d}$

3. 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...

2. Oct 31, 2016

### TSny

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, $dx$.

3. Oct 31, 2016

### RingNebula57

Thank you!

I solved it. Obtaining that :

$\Delta W = W \cdot (\eta-1)$

4. Oct 31, 2016

Looks good!