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

In summary: Your example of a discharging capacitor is a good example of this. The current defined by ##I=dq/dt## is negative because the charge flowing through the capacitor is decreasing, but the current as defined by ##I=\int \vec J \cdot \hat n~dA## is positive because the direction of the current density, ##\vec J##, is in the same direction as the chosen normal. In summary, the negative sign in front of ##dq/dt## is necessary to make the current, ##I##, positive and consistent with the chosen direction of the current density, ##\vec J##.
  • #1
What does the negative sign tell?
 
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  • #2
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.
 
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  • #3
Does it mean I has to be always kept positive?
 
  • #4
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
Zubair Ahmad said:
What does the negative sign tell?

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
Zubair Ahmad said:
Does it mean I has to be always kept positive?
No
 
  • #7
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.
 

1. What does the equation I = -dq/dt represent?

The equation I = -dq/dt represents the relationship between current (I) and charge (q) over time (t). It is known as the current-charge relationship and is used to describe the rate at which charge is flowing through a circuit.

2. Why is the negative sign included in the equation?

The negative sign in the equation represents the direction of current flow. It is included because current is defined as the flow of positive charge, but in reality, it is the movement of negative charges. Therefore, the negative sign ensures that the direction of current flow aligns with the direction of negative charge movement.

3. How is the equation I = -dq/dt used in circuits?

The equation I = -dq/dt is used to calculate the current at any given point in time. It is commonly used in circuits to determine the rate at which charge is being transferred through a component or along a wire. It can also be used to analyze the behavior of capacitors and inductors.

4. What is the relationship between charge and current in this equation?

In this equation, charge (q) and current (I) have an inverse relationship. This means that as charge decreases, current increases, and vice versa. This relationship is demonstrated by the negative sign in the equation, which indicates that as charge decreases (dq < 0), current will be positive (I > 0).

5. Is this equation always valid for all types of circuits?

No, this equation is only valid for circuits with constant current. In circuits with changing current, such as AC circuits, the equation becomes more complex. Additionally, this equation does not account for the resistance in a circuit, so it is not applicable in circuits with varying resistance.

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