How much energy in microjoules is stored in the capacitor

[ME_student]

Homework Statement

a) How much energy in microjoules is stored in the capacitor at t=500 micro sec
b) Repeat (a) for t=infinite

Homework Equations

Will be on attachment

The Attempt at a Solution

I am not sure how to approach the problem.. The graph shown is a graph of amps respect to time. So the tangent line to the function curve is joules?

 

[gneill]

The tangent to a current versus time curve would be ΔI/Δt, or di/dt. That's not energy.

What's a formula for the energy stored on a capacitor? You should know one in terms of voltage and another in terms of charge (Coulombs). How might you find the total charge or voltage on the capacitor from the given curve? Note that you also have the formula for the curve, so you can work the problem symbolically.

 

[ME_student]

Energy stored on a capacitor is given by C=(epsilon*A)/(d)

Epsilon is a constat
A is the area of the plate
d is the distance between the plates

the equation with voltage is Q=C*dv/dt

 

[ME_student]

I am currently working on the problem. I am trying to understand what the tangent to the graph or the total area underneath a graph represents when you have a graph that may have voltage respect to resistance, power respect to resistance, current respect to time, etc.

 

[ME_student]

I said Q=integral of 50e^-2000t with limits of integration of 0 to 500microsec. I get 0.015803mA then I used C=Q/v to get volate. After that I use w=1/2(CV^2)

But I don't get the correct answer.

 

[gneill]

Energy stored on a capacitor is given by C=(epsilon*A)/(d)

Epsilon is a constat
A is the area of the plate
d is the distance between the plates

You seem to be having some difficulty distinguishing the units of various electrical quantities. The above formula yields C, the Capacitance of the capacitor, not the energy. Energy has the units Joules.

the equation with voltage is Q=C*dv/dt

And that's an expression for the current (I in amps) , or rate of change of change charge (Coulombs) on the capacitor.

 

[gneill]

I am currently working on the problem. I am trying to understand what the tangent to the graph or the total area underneath a graph represents when you have a graph that may have voltage respect to resistance, power respect to resistance, current respect to time, etc.

If you integrate current over time what do you get? Remember that current is the rate of flow of charge, or dQ/dt. That means the area under the curve represents...?

 

[gneill]

I said Q=integral of 50e^-2000t with limits of integration of 0 to 500microsec. I get 0.015803mA then I used C=Q/v to get volate. After that I use w=1/2(CV^2)

But I don't get the correct answer.

Okay, that method is correct, but your value for charge Q is not correct and the units will not be current (mA) but Coulombs (C). Can you show more detail for your integration steps?

 

[ME_student]

I got it figured out... I used V(t)=1/C integral i dT+V(0) I got 11.06v and then used w=1/2*CV^2 to find energy