Collection of Solved Problems in Physics

Charge Transferred Through a Conductor

Task number: 1790

• Notation

  \(t = 10\,\mathrm{s}\)	Time during which the current flows through the conductor
  \(I = 5\,\mathrm{A}\)	Constant current
  \(I_1 = 0\,\mathrm{A}\)	Starting current
  \(I_2 = 3\,\mathrm{A}\)	Final current
  \(Q = ?\,\mathrm{[C]}\)	Transferred charge

• Hint I.

  What is the relation between current, charge and time?

• Hint solution

  The electric current I is defined as the ratio of the electrical charge Q transferred through the conductor over time t:

  \[I=\frac{Q}{t}.\]

• Solution I.

  For the constant current, we calculate the charge from the formula

  \[I=\frac{Q}{t}.\]

  The charge Q transferred by the current I over time t is then

  \[Q = I \cdot t = 5\,\mathrm{s} \,\cdot\, 10\,\mathrm{A}= 50\,\mathrm{C}.\]

• Hint II.

  The current is not constant in the second case. Try to draw the graph of current versus time and think if you can assess the transferred charge from this graph.

• Hint solution

  The current increases steadily in the second case. This means a linear dependence of current on time which can be plotted as the following graph.

  

  The formula for charge transferred by a constant current: \(Q = I \cdot t\) in fact represented the area of a rectangle with length t and height I. In our case, we can segment the graph of the increasing current into narrow rectangular stripes whose area we can calculate using the same formula. Let us segment the total time interval t into short intervals Δt. The current does not change much over one time interval — we can calculate the charge transferred over one time step again as \(Q = I(t) \cdot \Delta t\), where I(t) is the approximate current during the given time step. It is clear that the sum of the small charges transferred in individual time steps gives the total transferred charge.

  Geometrically, \(Q = I(t) \cdot \Delta t\) is the area of narrow vertical stripes with width Δt and with a height equal to the current at the given time — the sum of the areas of these stripes is euqal to the total area under the graph. It follows that the charge transferred by a current that changes over time in any way is equal to the area under the graph of the given current versus time.

  The area under our graph of current is the area of a right trapezoid (see the pink area in the figure below).

  

  We can calculate the area of the trapezoid simply when we notice that the two triangles highlighted with the blue line have the same area. The area of the trapezoid is thus equal to the area of a rectangle with a height in the middle of the rising part of the graph. We find that the charge transferred by a steadily (linearly) increasing current is the same as if the charge was transferred by a constant current I_ave whose value is the average of the initial value I₁ and the final value I₂ of the linearly increasing current. We can thus write for the transferred charge:

  \[Q=I_\mathrm{ave} \cdot t = \frac{I_1 + I_2}{2}t,\]

  where t is the time over which the current increases.

• Solution II.

  The total charge transferred by the steadily increasing current is:

  \[ Q=I_\mathrm{ave} \cdot t = \frac{I_1 + I_2}{2}t=\frac{0+3}{2}\cdot10= 15\,\mathrm C, \]

  where I_ave is the average of the inital current I₁ and the final current I₂.

• Comment – Solving by integration

  The task can be solved for any non-constant current by integration. The charge is in general given by the definite integral of current I(t) over time t from time t₁ to time t₂:

  \[Q=\int_{t_1}^{t_2} I(t)\,\mathrm dt.\]

  This integral in fact calculates the area under the graph of current versus time. The derivation in Hint Solution II. indicated a summation of the areas of vertical stripes with width Δt and height of the actual current in the given time interval I(t). Integration means nothing more than a very precise summation — with infinitely narrow time segments dt.

  We need to express the dependence of current on time I(t). We know that our dependence is linear, i.e. it is described in general by the formula: \[I=at+b,\] where a, b are constants which we assess from the assigned values. We substitute the starting and the final current into the formula and thus get a system of two linear equations from which we get these two constants. \[I_1=at_1+b\] \[I_2=at_2+b\] Using the assigned values in SI units we get the equations \[0=a\cdot 0+b,\] \[3=a\cdot 10+b .\] We can see directly from the first equation that b = 0. Substituting b into the second equation yields a = 0.3 . We thus have the formula of our linear function \[I=0.3t+0.\] And we calculate the integral (also in SI units)

  \[\{Q\}=\int_{0}^{10} 0.3t\,\mathrm dt=0.3\int_{0}^{10} t\,\mathrm dt=0.3\,\left[\frac{t^2}{2}\right]_0^{10}=0.3 \cdot \left[\frac{10^2-0^2}{2}\right]=15\] \[Q= 15\,\mathrm C.\]

• Proof that the average current can be used

  Let us now prove, for the sake of completeness, that integration also leads to the average value of current that was used as the principal solution above.

  We integrate a general linear dependence of current on time:

  \[Q=\int_{t_1}^{t_2} I(t)\, \mathrm dt=\int_{t_1}^{t_2} at+b\, \mathrm dt=\left[a\frac{t^2}{2}+bt \right]_{t_1}^{t_2}=a\frac{\left(t_2-t_1\right)^2}{2}+b\left(t_2-t_1\right)=\frac{a\left(t_2-t_1\right)^2+2b\left(t_2-t_1\right)}{2}\] \[Q=\frac{a\left(t_2-t_1\right)+2b}{2}\left(t_2-t_1\right).\]

  The elements in the denominator above can be grouped as:

  \[Q=\frac{a\left(t_2-t_1\right)+2b}{2}\left(t_2-t_1\right)=\frac{\left[at_1+b\right]+\left[at_2+b\right]}{2}\left(t_2-t_1\right). \]

  The brackets give exactly the definitions of the initial and the final current, respectively (see the previous comment) and the difference (t₂ − t₁) gives the total time over which we measure the transferred charge:

  \[Q=\frac{\left[at_1+b\right]+\left[at_2+b\right]}{2}\left(t_2-t_1\right)=\frac{I_1+I_2}{2}t\]

  We thus see that for a linear dependence of current on time, the total transferred charge is indeed equal to the charge transferred over the same time interval by a current whose value is the average of the initial and the final current.

• Answer

  1. A charge of \(50\, \mathrm C\) is transferred through a conductor by the current \(5\,\mathrm A\) over \(10\,\mathrm s\).
  2. A charge of \(15\,\mathrm C\) is transferred with the current that steadily increases from zero value to \(3\,\mathrm A\)over the same time.