# Velocity of capacitor plates

1. May 16, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
In the following RC circuit, the capacitor is in the steady state. The initial separation of the capacitor plates is x0. If at t = 0, the separation between the plates starts changing so that a constant current flows through R. Find the velocity of the moving plates as a function of time. The plate area is A.

3. The attempt at a solution

Applying Kirchoff's Law

$E=\dfrac{qvt}{\epsilon_0 A} + \dfrac{R dq}{dt}$

But how do I solve this differential equation?

#### Attached Files:

• ###### image050.gif
File size:
1.7 KB
Views:
63
2. May 16, 2014

### vela

Staff Emeritus
The differential equation you derived isn't quite correct because the plate separation is initially $x_0$, which doesn't appear anywhere in your equation.

After you fix that, the quantities that potentially vary with time are q, v, and dq/dt, which makes the equation hard to solve. You want to eliminate q and dq/dt and get an equation that only depends on v and t. Use the information that the current through the resistor is constant.

3. May 16, 2014

### utkarshakash

The correct equation should be

$E=\dfrac{q(x_0-\int vdt )}{\epsilon_0 A} + \dfrac{R dq}{dt}$

Here I've assumed that the separation is decreasing.

Last edited: May 16, 2014
4. May 17, 2014

### rude man

??
That equation is dimensionally inconsistent in the 1st term on the right.

Hint: what is the current thru the capacitor in terms of capacitance and capacitor voltage?
Combine that with vela's hint, i.e. E = V_C + V_R. You wind up with a simple ODE in C(t), without either V or q.

5. May 17, 2014

### utkarshakash

$q=CV_c \\ \dfrac{dq}{dt} = i=V_c \dfrac{dC}{dt} \\ V_r = iR \\ E= \dfrac{q}{C} + V_c R \dfrac{dC}{dt}$

Are the above equations correct?
Why is my earlier equation dimensionally inconsistent?

6. May 17, 2014

### rude man

EDIT:
They are correct but the fourth is unhelpful.
What can you say about Vc in terms of E, i0 and R where i0 is the constant current? What is i0 in this problem?

Also, get rid of q in your last equation.

Last edited: May 17, 2014
7. May 17, 2014

### utkarshakash

Vc = E-i0 R

8. May 17, 2014

### rude man

Excellent idea! Now, get rid of q.

9. May 17, 2014

### utkarshakash

The last equation can be modified as

$E = (E-i_0 R)(1+R\dfrac{dC}{dt})$

10. May 17, 2014

Yes!