Solved Problem on Electric Current
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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.
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).
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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