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Net and Total Charge from Current

Use signed and absolute-current integrals to distinguish net charge transfer from the total amount of charge that moves.

Advanced 9 min read

What You'll Learn

  • Interpret current as the rate of change of electric charge.
  • Calculate net charge transfer by integrating signed current over time.
  • Find total charge moved by integrating the absolute value of current.
  • Locate current sign changes and split an integral at the relevant times.
  • Use current units to report charge in coulombs.
  • Distinguish charge transferred from the resulting change in charge state.

Why Current Must Be Integrated

Electric current describes how quickly electric charge passes a chosen reference point. If the current remains constant, multiplying it by elapsed time gives the amount of charge transferred. When current varies, integration performs the corresponding accumulation over the time interval.

The direction of current matters. Choose one direction through the reference point as positive. Current in the opposite direction is negative under that convention. As a result, integrating current gives net charge transfer: transfer in the positive direction counts positively, and transfer in the opposite direction counts negatively. To find the total amount transferred in either direction, use the absolute value of current before integrating.

Definition: If \(Q(t)\) is charge, measured in coulombs, that has passed a reference point by time \(t\), then current is its rate of change: \(i(t)=Q'(t)=\frac{dQ}{dt}\). Current is measured in amperes, where \(1\text{ A}=1\text{ C/s}\).

Net Charge Transfer

Suppose the current \(i(t)\) is continuous, or piecewise continuous, on a time interval \([a,b]\), and \(Q'(t)=i(t)\). The net charge transferred during that interval is the change in accumulated charge:

$$ Q(b)-Q(a)=\int_a^b i(t)\,dt. $$

A positive integral means that the net transfer is in the chosen positive direction. A negative integral means that the net transfer is in the opposite direction. The integral can be zero even when substantial charge moved, if transfers in the two directions cancel.

The units confirm the interpretation: amperes multiplied by seconds give coulombs. If time is measured in seconds and current in amperes, the integral has units \(\text{A}\cdot\text{s}=\text{C}\).

Theorem: For a current \(i\) that is continuous or piecewise continuous on \([a,b]\), if \(Q'(t)=i(t)\), then the net charge transferred from \(a\) to \(b\) is \(\int_a^b i(t)\,dt\). This follows from the Fundamental Theorem of Calculus applied to \(Q\).

Total Charge Moved

Net transfer does not count all movement in the same way: opposite directions cancel. If a problem asks how much charge moved altogether, every direction should contribute a nonnegative amount. The rate of that total accumulation is therefore \(|i(t)|\), not \(i(t)\).

Definition: The total amount of charge moved through a reference point on \([a,b]\) is \(\int_a^b |i(t)|\,dt\). It is nonnegative and is measured in coulombs.

Because \(|i(t)|=i(t)\) when current is positive and \(|i(t)|=-i(t)\) when current is negative, the usual procedure is to find where current changes sign and split the interval there. A zero of \(i\) is a possible sign-change point, but not every zero changes the sign. Check the sign on each resulting interval.

Why the Two Integrals Give Different Answers

The Fundamental Theorem of Calculus explains the signed integral directly: because \(i=Q'\), integrating \(i\) measures the change in \(Q\). For total movement, consider a short time interval of length \(\Delta t\). If current is nearly constant on that interval, the amount of charge moved is approximately \(|i(t)|\Delta t\), regardless of direction. Adding these nonnegative amounts across the time interval leads to the integral of \(|i(t)|\).

This also explains why the absolute value must be applied to the current before integration. Taking the absolute value after integration only removes the sign of the net transfer; it cannot restore amounts that already canceled.

1
Identify the interval and units.
Use current in amperes and time in seconds, unless the problem specifies another consistent unit.
2
Find the current’s sign.
Locate zeros and any points where the current’s formula changes. Determine the sign on each interval.
3
Choose the requested quantity.
Integrate \(i(t)\) for net transfer, or integrate \(|i(t)|\) for total charge moved.
4
Evaluate and check.
Use signed contributions for net transfer and nonnegative contributions for total movement. Confirm the final unit is coulombs.

Worked Examples

Worked Example: A Current That Reverses Direction

Let \(i(t)=3t-6\) amperes for \(0\le t\le4\) seconds. Find the net charge transfer and the total charge moved.

First find when the current is zero:

$$ 3t-6=0 \quad\Longrightarrow\quad t=2. $$

For \(0\le t<2\), \(3t-6<0\); for \(2<t\le4\), \(3t-6>0\). An antiderivative is \(F(t)=\frac{3}{2}t^2-6t\).

The net transfer is the signed integral over the whole interval:

$$ \begin{aligned} \int_0^4(3t-6)\,dt &=F(4)-F(0)\\ &=\left(\frac{3}{2}(16)-24\right)-0\\ &=24-24\\ &=0\text{ C}. \end{aligned} $$

For total charge moved, split at \(t=2\) and make each contribution nonnegative:

$$ \begin{aligned} \int_0^4|3t-6|\,dt &=-\int_0^2(3t-6)\,dt+\int_2^4(3t-6)\,dt\\ &=-\bigl(F(2)-F(0)\bigr)+\bigl(F(4)-F(2)\bigr)\\ &=-\bigl((6-12)-0\bigr)+\bigl(0-(-6)\bigr)\\ &=6+6\\ &=12\text{ C}. \end{aligned} $$

Check: The first half transfers \(6\) C in the negative direction, and the second transfers \(6\) C in the positive direction. Their signed sum is \(0\) C, while their total is \(12\) C.

Worked Example: Piecewise-Constant Current

A current is \(4\) A from \(t=0\) to \(t=2\) s, \(-3\) A from \(t=2\) to \(t=5\) s, and \(2\) A from \(t=5\) to \(t=6\) s. Find net transfer and total charge moved over the six seconds.

For a constant current, the integral on each interval is current multiplied by interval length. The three signed contributions are:

$$ \begin{aligned} (4\text{ A})(2\text{ s})&=8\text{ C},\\ (-3\text{ A})(3\text{ s})&=-9\text{ C},\\ (2\text{ A})(1\text{ s})&=2\text{ C}. \end{aligned} $$

Thus the net transfer is

$$ 8+(-9)+2=1\text{ C}. $$

For total charge moved, take the magnitude of each interval’s contribution:

$$ |8|+|-9|+|2|=8+9+2=19\text{ C}. $$

Check: The positive-direction amounts total \(8+2=10\) C, and the opposite-direction amount is \(9\) C. Their net is \(10-9=1\) C, and their combined total is \(10+9=19\) C.

Worked Example: A Quadratic Current and an Initial Charge

Suppose \(i(t)=t^2-4t+3\) amperes for \(0\le t\le4\) seconds, and the initial charge value is \(Q(0)=10\) C. Find the net transfer, the total amount moved, and \(Q(4)\).

Factor the current to find its zeros:

$$ i(t)=(t-1)(t-3). $$

The current is positive on \([0,1)\), negative on \((1,3)\), and positive on \((3,4]\). An antiderivative is \(F(t)=\frac{t^3}{3}-2t^2+3t\). Its values at the needed endpoints are

$$ F(0)=0,\qquad F(1)=\frac13-2+3=\frac43,\qquad F(3)=9-18+9=0,\qquad F(4)=\frac{64}{3}-32+12=\frac43. $$

The net transfer is

$$ \int_0^4 i(t)\,dt=F(4)-F(0)=\frac43\text{ C}. $$

The total charge moved is the sum of the magnitudes on the three sign intervals:

$$ \begin{aligned} \int_0^4|i(t)|\,dt &=\bigl(F(1)-F(0)\bigr)-\bigl(F(3)-F(1)\bigr)+\bigl(F(4)-F(3)\bigr)\\ &=\frac43-\left(0-\frac43\right)+\frac43\\ &=\frac43+\frac43+\frac43\\ &=4\text{ C}. \end{aligned} $$

Since the net transfer is the change in charge, \(Q(4)=Q(0)+\frac43=10+\frac43=\frac{34}{3}\) C. Check: Subtracting the initial value gives \(\frac{34}{3}-10=\frac43\) C, exactly the net integral. The total moved, \(4\) C, is larger because it counts the reversal in direction without cancellation.

A Common Mistake

A frequent error is to compute the net integral and then take its absolute value, assuming that this gives total charge moved. For the first example, that incorrect calculation would be

$$ \left|\int_0^4(3t-6)\,dt\right|=|0|=0\text{ C}. $$

This answer is the magnitude of net transfer, not total movement. The current carries \(6\) C in one direction and \(6\) C in the other. The correction is to take the absolute value before integrating:

$$ \int_0^4|3t-6|\,dt=12\text{ C}. $$
Key takeaway: Use \(\int_a^b i(t)\,dt\) for net charge transfer. Use \(\int_a^b|i(t)|\,dt\) for total charge moved. If the current changes sign, split the total-charge calculation at its sign changes.

Check Your Understanding

For each question, distinguish signed net transfer from total charge moved when both are requested. Assume time is measured in seconds and current in amperes.

  1. If \(i(t)=5\) on \(0\le t\le3\), what are the net transfer and total charge moved?
  2. If \(i(t)=-2\) on \(1\le t\le4\), what are the net transfer and total charge moved?
  3. For \(i(t)=t-2\) on \(0\le t\le4\), at what time does the current change sign, and what are the net transfer and total charge moved?
  4. Why is \(\left|\int_a^b i(t)\,dt\right|\) not generally equal to \(\int_a^b|i(t)|\,dt\)?
  5. If \(Q(0)=7\) C and the net transfer from \(0\) to \(5\) seconds is \(-3\) C, what is \(Q(5)\)?

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