What is the Velocity of the Capacitor Plates in an RC Circuit?

[utkarshakash]

Homework Statement

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.

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?

 

[vela]

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.

 

[utkarshakash]

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.

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.

 

[rude man]

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.

??
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.

 

[utkarshakash]

??
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.

q=CV_c \\<br /> \dfrac{dq}{dt} = i=V_c \dfrac{dC}{dt} \\<br /> V_r = iR \\<br /> E= \dfrac{q}{C} + V_c R \dfrac{dC}{dt}

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

 

[rude man]

EDIT:
They are correct but the fourth is unhelpful.
What can you say about V_c 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.

 

[utkarshakash]

EDIT:
They are correct but the fourth is unhelpful.
What can you say about V_c 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.

Vc = E-i₀ R

 

[rude man]

Vc = E-i₀ R

Excellent idea! Now, get rid of q.

 

[utkarshakash]

Excellent idea! Now, get rid of q.

The last equation can be modified as

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

 

[rude man]

The last equation can be modified as

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

Yes!