Time constant of a discharge RC circuit, capacity and charge

[greg_rack]

Homework Statement

A charged capacitor of potential $V=20V$ is connected to a resistance $R=1k\Omega$.
While discharging, at time $t_{1}=0.12s$ the charge is $q_{1}=40\mu C$, whilst at time $t_{2}=0.20s$ the charge is $q_{2}=5.4\mu C$. Determine:
-the time constant of the circuit;
-the charge on the capacitor at instant $t_{0}=0s$;
-the capacity of the capacitor.

Relevant Equations

$q=CV$
Discharge function: $q(t)=q_{0}e^{-\frac{t}{\tau}}$

So, the only thing which came to my mind in order to solve this problem was actually to write down the equations using the discharge function, being given two instants and their corresponding charges... but doing so I'm unable to find anything.
Ideally, I'd say I should find the time constant $\tau$, then the capacity, and lastly the maximum charge at the beginning of the process.

 

[etotheipi]

It's actually easier to shift your zero of time to $t' = t- 0.12s$ and then write the discharge equation with initial charge $q_1$, but to see this more clearly you can start by writing:$$\begin{align*}q_1 &= q_0 e^{-\frac{t_1}{\tau}} \\ q_2 &= q_0 e^{-\frac{t_2}{\tau}}\end{align*}$$What do you get if you divide these two?

 

[greg_rack]

It's actually easier to shift your zero of time to $t' = t- 0.12s$ and then write the discharge equation with initial charge $q_1$, but to see this more clearly you can start by writing:$$\begin{align*}q_1 &= q_0 e^{-\frac{t_1}{\tau}} \\ q_2 &= q_0 e^{-\frac{t_2}{\tau}}\end{align*}$$What do you get if you divide these two?

What do you mean by "shifting my zero to $t' = t- 0.12s$"? What's $t$?
Actually, by equalling the $q_{0}$ from both expressions, I get to solve an exponential equation with just one unknown, which is $\tau$... but that's a pretty tricky one!
There must be some easier way compared to solving an expo.

 

[etotheipi]

If you divide those two equations, you get$$q_2 = q_1 \text{exp}({-\frac{t_2 - t_1}{\tau}})$$You can solve that for $\tau$ by taking $\ln(\cdot)$ of both sides.

What I was saying about the shifting your origin of time was that, if you define $t' = t - t_1$, then the capacitor has charge $q_1$ at $t' = 0$ and $q_2$ at $t' = t_2 - t_1$. Then, you can just write down the equation above! But this isn't so important, so long as you can get the equation one way or another.

 

[greg_rack]

If you divide those two equations, you get$$q_2 = q_1 \text{exp}({-\frac{t_2 - t_1}{\tau}})$$You can solve that for $\tau$ by taking $\ln(\cdot)$ of both sides.

What I was saying about the shifting your origin of time was that, if you define $t' = t - t_1$, then the capacitor has charge $q_1$ at $t' = 0$ and $q_2$ at $t' = t_2 - t_1$. Then, you can just write down the equation above! But this isn't so important, so long as you can get the equation one way or another.

Yeah, that's right! Cool!
I'm feeling quite dumb asking you this question, but: in a system, am I always legitimized to divide the equations to obtain an equivalent one?

 

[etotheipi]

Yeah, that's right! Cool!
I'm feeling quite dumb asking you this question, but: in a system, am I always legitimized to divide the equations to obtain an equivalent one?

Yes! - well with the usual stipulation that you're not dividing by zero. Say you have two equations,$$\begin{align*}
a&=b \\
c&=d
\end{align*}$$Divide the first equation by $c$, with $c\neq 0$, to get$$\frac{a}{c} = \frac{b}{c}$$
But $c=d$, so$$\frac{a}{c} = \frac{b}{d}$$

 

[greg_rack]

Yes! - well with the usual stipulation that you're not dividing by zero. Say you have two equations,$$\begin{align*}
a&=b \\
c&=d
\end{align*}$$Divide the first equation by $c$, with $c\neq 0$, to get$$\frac{a}{c} = \frac{b}{c}$$
But $c=d$, so$$\frac{a}{c} = \frac{b}{d}$$

Thank you so much