Rates of Change in a RC Circuit

[Thefox14]

Homework Statement

A 2.9 MΩ and a 2.8 μF capacitor are connected in series with an ideal battery with an EMF of 3 V. At 2 seconds after the circuit is initially connected to the battery:

What is the rate at which energy is being stored on the capacitor?

Homework Equations

dq/dt = v/r * e^(-t/RC) = 8.08e-7

The Attempt at a Solution

I have tried differentiating .5Q^2/C and got (dq/dt)/C But this didnt work. I did the same for .5CV^2

What am I doing wrong?

 

[Andrew Mason]

Homework Statement

A 2.9 MΩ and a 2.8 μF capacitor are connected in series with an ideal battery with an EMF of 3 V. At 2 seconds after the circuit is initially connected to the battery:

What is the rate at which energy is being stored on the capacitor?

Homework Equations

dq/dt = v/r * e^(-t/RC) = 8.08e-7

The Attempt at a Solution

I have tried differentiating .5Q^2/C and got (dq/dt)/C But this didnt work. I did the same for .5CV^2

What am I doing wrong?

You have calculated the current at t = 2 sec. That current is working against the potential difference across the capacitor. What is that potential difference at this time? What is the relationship between current, voltage and power?

AM

 

[cupid.callin]

E = .5q²/C

$$\frac{dE}{dt}$$ = $$\frac{q}{C}$$$$\frac{dq}{dt}$$


$$\frac{dE}{dt}$$ = $$\frac{q}{C}$$$$i$$

so now for ant time t you can find q on capacitor and current in curcuit ...just substitute them

 

[Thefox14]

E = .5q²/C

$$\frac{dE}{dt}$$ = $$\frac{q}{C}$$$$\frac{dq}{dt}$$


$$\frac{dE}{dt}$$ = $$\frac{q}{C}$$$$i$$

so now for ant time t you can find q on capacitor and current in curcuit ...just substitute them

Ah, Thats right its the chain rule! I kept leaving out the q term heh.

Thanks I get the correct answer now.

 

[Andrew Mason]

A slightly different approach would be: the work done in moving a charge dq through a potential V is Vdq. So the rate of energy storage is Vdq/dt = VI where V = Q/C

AM