# Voltage across a capacitor variable of integration

I understand that something that is charged with one coulomb and has a potential difference of one volt across it will have a capacitance of one farad. $C = \frac{q}{V}$. I get that because current is the derivative of charge with respect to time, $I = C\frac{dV}{dt}$.

This is where I start to get confused. If you divide both sides by $C$ and integrate with respect to time, shouldn't that just give you $V = \frac{q}{C}$?

What I don't get is how this becomes $V(t) = \frac{1}{C}\int_{t_{0}}^{t}I(\tau)\, d\tau + V(t_{0})$.

This is saying that $q = \int_{t_{0}}^{t}I(\tau)\, d\tau + V(t_{0})$. I don't get why, though.

Shouldn't $q = \int I\, dt$ and that be the end of it? I mean, obviously this isn't the case since my textbook and everything I can find online says otherwise, but I don't get it.

q is not equal to $\int_{t0}^{t} I(\tau) d\tau + V(t_0)$

it has no sense to sum apples (charge) and pears (potential) :)

if you have $V(t)=\frac{1}{C}\int_{t0}^{t} I(\tau) d\tau + V(t_0)$

$\frac{1}{C}$ is just multiplying $\int_{t0}^{t} I(\tau) d\tau$ and not $V(t_0)$

$\int_{t0}^{t} I(\tau) d\tau$ is a $\Delta Q$

so $V(t)=\frac{1}{C}\int_{t0}^{t} I(\tau) d\tau + V(t_0)=\frac{\Delta Q}{C} +V(t_0)$

I hope this is what you were missing :)

This is really just a math question.

When you take the integral $\frac{1}{C}\int_{t_0}^t I(\tau)d\tau$, you get $\frac{Q(t)}{C} - \frac{Q(t_0)}{C}$ = $V(t) - V(t_0)$. This is bad, because we just want $V(t)$; the simple way of fixing this is to add $V(t_0)$ to cancel out the term in the integral.

We definitely can't just write $\frac{1}{C} \int I(t)$ - what are you going to do with the constant of integration?