# I If charge is decreasing with time then we write I =-dq/dt.

1. May 17, 2018

What does the negative sign tell?

2. May 17, 2018

### kuruman

It says that as time increases, quantity q decreases if I is a positive quantity. For example, if you have a capacitor discharging through a resistor in an RC circuit, you would write Ohm's Law as VR = I R, where I is the current through the resistor and by definition a positive number. If you wanted to relate that current I to the rate of change of charge q on the capacitor plates, you would write I = -(dq/dt). That's because the capacitor is discharging which means that dq/dt is a negative quantity therefore you need the negative sign in front to make the current I on the left side positive.

3. May 17, 2018

Does it mean I has to be always kept positive?

4. May 17, 2018

### kuruman

What I am saying is that the current I that appears in Ohm's Law, $V = IR$, has to be always positive. So if someone tells you that the charge on a capacitor is given by $Q(t)=Q_0e^{-t/(RC)}$ and asks you to find the current $I$ in the resistor, you would say $I=-(dQ/dt)$. This appears to be in contradiction to the definition of current that you see in textbooks, $I=dq/dt$, but it isn't.

5. May 17, 2018

### ZapperZ

Staff Emeritus
This is perfectly VAGUE! You need to make some effort in describing the situation.

Current is defined as the rate of charge flow across a cross-sectional surface. If the charge passing through per unit time is decreasing, dq/dt is still positive, but dI/dt is negative! This means that current is decreasing over time.

dq/dt will have a negative value if (i) q is a negative charge OR (ii) the positive charge is moving in the OPPOSITE direction. This implies that current is in the opposite direction.

Zz.

6. May 17, 2018

### Staff: Mentor

No

7. May 18, 2018

### kuruman

Here is my thinking on the matter. The total current $I$ through a surface is formally defined using the current density, $I=\int_S \vec J\cdot \hat n~dA$ where $\hat n$ is a chosen normal to the surface and such that $\vec J \cdot \hat n$ is positive, i.e. the charge carriers are moving (mostly) in the direction of $\vec J$. I think of current $I$ as a positive scalar, analogous to pressure $p=\frac{\vec F \cdot \hat n }{dA}$ where the force is the vector quantity that specifies direction. When current (or pressure) turn out to be negative, all this means is that the explicitly or implicitly chosen normal to the area element $dA$ forms an angle greater than $90^o$ relative to the current density (or force).

In the standard justification for $I=dq/dt$, one imagines amount of charge $dq$ moving by a fixed point in time $dt$. In that respect, current $I$ is like a one-dimensional speed $v=ds/dt$. If a one-dimensional speed (for whatever reason) turns out to be negative, we say that the velocity is "opposite" to the assumed direction; we never say that the speed is in the opposite direction. Likewise, it seems to me that, for consistency and to avoid confusing the vector with its magnitude, when the current $I$ or $dq/dt$ turn out to be negative, it is appropriate to think of the current density (or direction of carrier flow) being opposite to the assumed direction.

I prefer to think of current direction in terms of $\vec J$ in which case the sign of the charge carriers does not matter. It helps me keep negative signs sorted out.