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I'm solving a differential equation of the form

$$200q'+\frac{200}{3}q=3U_o-6U_6+3U_{12}$$ Where $U$ is the Heaviside step function.

$q(0)=0 [Coulombs] $

After solving it, I get that at time $t=18$ the charge on my capacitor is negative.

My teacher says this is interpreted as "the capacitor is discharging". But I don't think this is true because that information would only come from the derivative.

What does negative charge on a capacitor even mean then?

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What does negative charge on a capacitor even mean then?

According to the fundamental relationship

$$Q = CV$$

negative $Q$ implies negative $V$. There's nothing mysterious about this since, in any problem, one must choose a reference polarity for the capacitor voltage $V$ (practically, this is essentially the choice of which capacitor terminal you place the red lead of your voltmeter on).

The clearest way to determine if a capacitor is charging or discharging is to see if the energy is increasing (charging) or decreasing (discharging).

For a capacitor, the energy in terms of the charge $Q$ is

$$W_C = \frac{Q^2}{2C}$$

The time derivative is

$$\frac{dW_C}{dt} = p_C = \frac{Q}{C}\cdot\frac{dQ}{dt} = v_C\cdot i_C$$

The right-most term assumes the passive sign convention so positive (negative) power implies charging (discharging).

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