Why is the current in a conductor equal to zero when not connected to a circuit?

[tellmesomething]

Homework Statement

"As there is a large number of free electrons in random directions, the number of electrons crossing an area $\delta S$ from one side very nearly equals the number crossing from the other side in any given time interval. The electric current through the area is therefore 0"

Relevant Equations

None

Can someone provide me with a justification on this? As fair as I know current through an area is the rate of transfer of charge from one side of an area to the other.
So how does current become 0 in a conductor (not connected to a circuit just a plain conductor) just cause "the number of electrons crossing an area $\delta S$ from one side very nearly equals the number crossing from the other side in any given time interval. " I dont think I understand this very well.

 

[jbriggs444]

Can someone provide me with a justification on this? As fair as I know current through an area is the rate of transfer of charge from one side of an area to the other.
So how does current become 0 in a conductor (not connected to a circuit just a plain conductor) just cause "the number of electrons crossing an area $\delta S$ from one side very nearly equals the number crossing from the other side in any given time interval. " I dont think I understand this very well.

If there were a significant net flow of charge in one direction then what would happen to the charge distribution as a result? What would then happen to the net flow of charge?

You might also do some Googling about thermal noise.

The second law of thermodynamics has something to say as well.

 

[Steve4Physics]

Relevant Equations: None

... As fair as I know current through an area is the rate of transfer of charge from one side of an area to the other.

If
$N$ electrons per second cross area $\delta S$ moving left to right,
and simultaneously
$N$ electrons per second cross the same area moving right to left
then the net rate of charge flow is zero. The current is zero because the two opposite flows of charge cancel, resulting in no overall flow of charge.

Edit. @tellmesomething, just in case of possible confusion, note that the area $\delta S$ is an area inside the conductor - not an end. So $\delta S$ has metal on both sides,

 

[tellmesomething]

Relevant Equations: None

If
$N$ electrons per second cross area $\delta S$ moving left to right,
and simultaneously
$N$ electrons per second cross the same area moving right to left
then the net rate of charge flow is zero. The current is zero because the two opposite flows of charge cancel, resulting in no overall flow of charge.

Edit. @tellmesomething, just in case of possible confusion, note that the area $\delta S$ is an area inside the conductor - not an end. So $\delta S$ has metal on both sides,

Thata what I dont get, how do they cancel out, the formula is $i=\frac{dq}{dt}$
I dont see how the signs would change. .for each of the cases you described

 

[tellmesomething]

If there were a significant net flow of charge in one direction then what would happen to the charge distribution as a result? What would then happen to the net flow of charge?

You might also do some Googling about thermal noise.

The second law of thermodynamics has something to say as well.

I honestly don't know how the charge distribution would change? Hints?

 

[jbriggs444]

I honestly don't know how the charge distribution would change? Hints?

If you have charge moving from one side to the other, the other side will have more charge.

 

[tellmesomething]

If you have charge moving from one side to the other, the other side will have more charge.

Right and this is when there's no cell connected to give a continuous flow of charges yes so one side would have more charge...
So are you implying this shouldnt be possible ?

 

[jbriggs444]

Right and this is when there's no cell connected to give a continuous flow of charges yes so one side would have more charge...
So are you implying this shouldnt be possible ?

What happens if you cram lots of positive charge on one side and lots of negative charge on the other?

 

[tellmesomething]

What happens if you cram lots of positive charge on one side and lots of negative charge on the other?

A potential difference appears

 

[Steve4Physics]

Thata what I dont get, how do they cancel out, the formula is $i=\frac{dq}{dt}$
I dont see how the signs would change. .for each of the cases you described

$\frac{dq}{dt}$ is positive for charges moving (say) left to right.
$\frac{dq}{dt}$ is negative for charges moving right to left.
These two flows occur simultaneously.
When you add the positve and negative values, they cancel.

 

[tellmesomething]

$\frac{dq}{dt}$ is positive for charges moving (say) left to right.
$\frac{dq}{dt}$ is negative for charges moving right to left.
These two flows occur simultaneously.
When you add the positve and negative values, they cancel.

Oh okay. Thankyou

 

[jbriggs444]

A potential difference appears

Does this difference assist or prevent any further movement of charge?

 

[tellmesomething]

Does this difference assist or prevent any further movement of charge?

Assists

 

[DaveC426913]

Assists

So if one side of the conductor builds up a charge, it will assist in building up more charge on that side? Do we get a runaway effect?