Solved Problem on Electric Current

A current of 2 A flows in a wire, knowing that the elementary charge is 1.6 × 10−19. How many electrons pass through a cross-section of the wire during a time interval of 1 s?

 

Problem data:

  • Electric current:    i = 2 A;
  • Elementary charge:    e = 1.6 × 10−19 C;
  • Time interval:    Δt = 1 s.

Problem diagram:

The electrons move from left to right across a cross-section, highlighted in gray, the conventional current, i, is indicated in the direction opposite to the movement of the electrons (Figure 1).
Figure 1

Solution:

The electric current is the rate at which charge flows through the cross-section

\[ \begin{gather} \bbox[#99CCFF,10px] {i=\frac{\Delta q}{\Delta t}} \tag{I} \end{gather} \]

The amount of charge that crosses a given cross-section is the product of the number of charges that cross the section by the elementary charge

\[ \begin{gather} \bbox[#99CCFF,10px] {\Delta q=ne} \tag{II} \end{gather} \]

substituting the equation (II) into equation (I)

\[ \begin{gather} i=\frac{ne}{\Delta t}\\[5pt] n=\frac{i\;\Delta t}{e}\\[5pt] n=\frac{\left(2\;\mathrm A\right)\left(1\;\mathrm s\right)}{1.6\times 10^{-19}\;\mathrm C}\\[5pt] n=\frac{2\times 10^{19}\;\cancel{\mathrm C}}{1.6\;\cancel{\mathrm C}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {n=1.25\times 10^{19}\;\text{electrons}} \end{gather} \]
Note: the electron charge is equal to −1,6×10−19 C. Since we want the number of electrons, n, to be positive, we use the positive value for the electron charge to ensure that the result is not negative.
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