# Why doesn't current flow when the wire is open?

I realize this may be a hard to answer question but we are currently studying current in our school. One thing that struck me was why the heck doesn't it flow when the wire's closed?

When you connect the positive terminal of the battery (But do not connect the negative terminal) to a conductor, the loosely bound / free electrons in the conductor should be attracted to the battery's +ve terminal. This should create a chain reaction till every atom in the conductor has donated the maximum it can and no further flow is possible due to the atoms now strongly attracting their remaining electrons.

Why doesn't this happen? There should be momentary current and the wire should become positively charged. Why not? Why does it only flow when the circuit is closed?

• There is a step current (just no steady-state current) as electrons redistribute to lowest energy in the new electric field but it generally lasts a very short time. – xish Jun 21 '14 at 2:14
• You know that if you add a resistance in series the current goes down, right? Then think of it this way: the resistance of air (it's dielectric constant) is really high. So, by opening a switch or disconnecting wires you are adding a huge resistance to the circuit (think Teraohms). This effectively reduces the current to zero. – HardScale Jun 12 '16 at 11:37

At first, try to understand the working principle of a battery. Here is a video explaining what positive terminal and negative terminal are. If you don't have the load connected there, you will not have infinite amount of depletion of electrons in the positive terminal as after some time no $H^+$ will be able to reach the positive terminal due to its electric field. similar statement for negative terminal.
so in the open circuit condition If you connect the conductor with the positive terminal only few of the electrons will go to that $PbO_2$ bar. Now the $H^+$ ions can again reach to the $PbO_2$, but it will sustain only for tiny amount of time as again $H^+$ will not be able to reach the positive terminal due to the electric filed of the terminal. Now this time $PbO_2$ bar+ the conductor becomes the positive terminal.