# How to compute the charge as a function of time in a simple $CL$ circuit?

I am trying to compute $q(t)$ in a series circuit consisting of a capacitor of plate charges $q_0$ and $-q_0$ and an inductor. The switch is closed at time $t=0$.

By Kirchoff's Law, $$\frac{q}{C}+L\frac{dI}{dt}=0.$$ There is a conservation of charge in this circuit, so $I=\dot{q}$, which gives us the differential equation $$\ddot{q}+\frac{1}{LC}q=0.$$ I recognise that this is the equation of a harmonic oscillator with angular frequency $\frac{1}{\sqrt{LC}}$, so $$\ddot{q}+\omega^2 q=0.$$ To solve this I construct a characteristic equation $$\lambda^2+\omega^2=0 \implies \lambda=\pm i\omega$$ and taking only the positive root gives the solution $q(t)$ as $$q(t)=A\cos{\omega t}+Bi\sin{\omega t}.$$ The total initial charge in the circuit is from the capacitor and is given by $q(0)=CV$, so we get $A\cos{\omega t}=CV \implies A=CV.$ Then $$q(t)=CV\cos(\omega t)+Bi\sin(\omega t).$$ I know that the answer should not have the imaginary term in it, but how to I eliminate it or show that B=0?

• This is a second order differential equation, so it's only well defined if you have two boundary conditions. You listed one: $q(0)=CV$, but there must be another. – DanielSank Dec 18 '17 at 21:20
• Hint: What is the current at time $t=0$?, or, more explicitly, $I(0)=\dot{q}(0)$. – jim Jun 28 '18 at 8:53

Recognize the initial condition: at t=0, no current is flowing. That means that the charge is a maximum at t=0, and that means that B=0. For a different initial condition, the equation just means there is a phase shift - that is, you could write $\cos\omega t + \phi$ instead of introducing a complex amplitude.
When we use complex numbers for solving harmonic equations, it is really just because it usually makes the math easier (more so if you use $Ae^{i\omega t}$ notation with $A$ a complex number): the actual amplitude at a given moment in time is the real part of the expression.