# Graph of energy stored in capacitor?

Question: In an experiment a capacitor is discharged through a constant current. Draw a graph of how the energy stored in the capacitor varies with time.

The answer given is:

But I seemed to get a different answer:

I know my graph is counter-intuitive since if you are discharging a capacitor it has to start off with a nonzero energy stored, so my graph being 0 at t=0 is contradictory. However, my working seemed to show that my answer is correct, so I want to know where I went wrong.

Basically for constant current we have $$Q=It$$ proportional to $$t$$, i.e. $$Q$$ proportional to $$t$$. Then since $$V$$ proportional to $$Q$$, we must have: $$V$$ proportional to $$Q$$ proportional to $$t$$, i.e. $$V$$ proportional to $$t$$. Therefore, the energy stored, which is given by $$E=\frac{1}{2}QV$$, is proportional to $$t^2$$, This means that $$E=kt^2$$ for some constant $$k$$. So it is a parabola like the one I showed.

Where did I go wrong?

• The equation $Q=It$ is incorrect, since $Q$ should be nonzero when $t=0$. Commented May 3, 2015 at 22:32
• @leongz: It must be, thanks! If you write an answer I'll accept quickly. (Please also include how the correct graph is found if you decide to write answer) Commented May 3, 2015 at 22:33
• Alternatively, you can write up your own answer, and show us how you can get correct graph. Commented May 3, 2015 at 22:37
• Try starting with $Q=Q_0-It$, where $Q_0$ is the initial charge the capacitor has. Commented May 3, 2015 at 22:41
• @leongz: I finished the answer, and thank you again for helping me understand Commented May 3, 2015 at 23:22

I understand the problem now from leongz's excellent explanation and patience in the chat discussion. Thank you very much leongz for helping me out!

The capacitor is being discharged through constant current. If it starts with charge $$Q_0$$, then from the definition of current we know that the charge decreases by It after time t, where I is the constant current. So $$Q=Q_0-It$$.

We want the graph of energy, so we use $$E=0.5\times Q\times V$$. Then we can eliminate V by using $$C=Q/V$$ to get $$V=Q/C$$. Putting this into $$E=0.5\times Q\times V$$ gives $$E=(0.5/C)(Q^2)$$.

But from before, $$Q=Q_0-It$$, so $$E=(0.5/C)(Q_0-It)^2$$.

We see that:

At $$t=0$$ this gives $$E=(0.5/C)(Q_0)^2$$, which is a positive $$y$$-intercept.

At $$E=0$$ we must have $$Q_0-It=0$$ which gives $$t=Q_0/I$$, which is positive $$x$$-intercept.

And also the graph is parabolic since the variable $$t$$ has degree 2.

Combining this we see that the graph is like the one in the answer.