# Voltage Current and Voltage Charge Relationship for a Capacitor

1. The voltage-current relationship for a capacitor is described by Eq. \eqref{1}: \begin{equation} \mathrm{v}\left(\, t\, \right) = \frac{1}{C}\int_{0}^{1}\mathrm{i}\left(\, \tau\, \right)\,\mathrm{d}\tau \label{1}\tag{1} \end{equation}
2. The voltage-charge relationship for a capacitor is described by Eq. \eqref{2}: \begin{equation} \mathrm{v}\left(\, t\, \right) = \frac{1}{C}\,\mathrm{q}\left(\, t\, \right) \label{2}\tag{2} \end{equation}
3. How can I prove that $\,\mathrm{q}\left(\, t\, \right) = \int_{0}^{1}\mathrm{i}\left(\, \tau\, \right)\,\mathrm{d}\tau$ ?.

The equation you want is $$q(t)= \int_0^t i(\tau) d\tau$$ This comes from integrating both sides of the definition $$\frac{dq}{dt}= i(t)$$
• Before performing the integration of both sides do I need to multiply both sides by $dt$ to get $dq = i(t)dt$? – SpaceBeer Aug 28 '16 at 6:13
• @SpaceBeer Both "yes" and "no" are possible here. You can do it without multiplying by $dt$, just integrating both sides as functions of $t$ and applying Newton-Leibnitz formula $\int \limits_{a}^{b} F'(x) dx = F(b) - F(a)$ to the $\frac{dq}{dt}$. – Evgeny Aug 28 '16 at 9:24
You can use your equations $(1)$ and $(2)$ to answer your question. From equation $(1)$, you have $$v(t) = \frac{1}C\int_0^1 i(\tau)\,d\tau\label{a}\tag{1}.$$
From your equation $(2)$, you have $$v(t) = \frac{1}{C}q(t)\label{b}\tag{2}.$$
Therefore, multiply equation $(2)$ by the capacitance $C$ on each side to obtain $$q(t) = Cv(t)\label{c}\tag{3}.$$ Substitute the value of $v(t)$ from equation $(1)$ into $(3)$, and you're done: $$q(t) = C\left(\frac{1}C\int_0^1 i(\tau)\,d\tau\right) = \int_0^1 i(\tau)\,d\tau.$$