Electric current
Slower than a snail
- Flip a switch and the light comes on instantly.
- Yet the electrons themselves drift through the wire slower than a snail.
- The push (the electric field) travels fast; the charges crawl.
What current is
- An electric current is a flow of charge carriers (electrons in a metal, ions in a liquid).
- Conventional current points the way positive charge would flow — opposite to the electrons.
Conventional current points in the direction that:
Conventional current is the flow of positive charge — opposite to the electron drift in a metal wire.
Charge comes in lumps
- Charge is quantised: the smallest free unit is the elementary charge $e = 1.60 \times 10^{-19}\ \text{C}$.
- Every free charge is a whole-number multiple of $e$. Unit of charge: the coulomb (C).
The smallest free unit of charge is the ____ charge, $e = 1.6 \times 10^{-19}\ \text{C}$.
All free charges are whole-number multiples of the elementary charge $e$.
Current = charge per second
- $I = \dfrac{Q}{t}$, so $Q = It$. Unit: the ampere ($1\ \text{A} = 1\ \dfrac{\text{C}}{\text{s}}$).
- For a changing current, the charge is the area under an $I$–$t$ graph.
A current of $2.0\ \text{A}$ flows for $5.0\ \text{s}$. How much charge passes?
$Q = It = 2.0 \times 5.0 = 10\ \text{C}$.
The charge that has flowed equals the area under an $I$–$t$ graph.
Yes — current is the rate of flow of charge, so the area (current × time) gives the total charge.
Drift velocity
- $I = Anvq$ — area $A$, carrier density $n$, drift speed $v$, charge $q$ each.
- Same current in a thinner wire → faster drift; a semiconductor (small $n$) → much faster drift.
Which equation gives the current in terms of the drift velocity?
Current = (area)(number density)(drift speed)(charge per carrier) $= Anvq$.
In a thinner wire carrying the same current, the electrons drift faster.
From $I = Anvq$, a smaller area $A$ at the same $I$ needs a larger drift speed $v$.
You've got it
- current is a flow of charge; conventional current = direction of positive flow
- $I = \dfrac{Q}{t}$, and charge is the area under an $I$–$t$ graph
- drift: $I = Anvq$ (charge comes in units of $e = 1.6 \times 10^{-19}\ \text{C}$)