If you connect some copper wire to a 9V battery terminal ( but don't make a circuit ), we say the wire now has a PD of 9V compared to the other battery terminal. But does it look any different to it's 0V self under nanoscopic inspection?
To answer this I use the water pressure analogy.
- 2 identical volumes of water.
- different pressures.
-> The one with higher pressure than the other has more water, mass , density.
-> Similarly, the wire at 9V has more charge in it
Me trying to answer :
Everything is a capacitor
The wire is a capacitor
When attached to the 9V terminal, it'll charge up
Also the other terminal is a capacitor, and it's charge will adjust according to the new capacitance of the 9V terminal and connected wire.
Use VC = q
So I reckon yes, the wire is physically and detectably different from its 0V self.
Also, in theory its possible to discharge\ use up a battery just by altering the capacitance of the terminals by connecting random items, e,g, a piece of wire, your finger, a coin, to one terminal but leaving the circuit open.
It would take a long time because the capacitance of such objects with the other terminal is so low hence a low charge hence a low energy usage to charge.
e.g.
I'm guessing by touching the 9V terminal with your finger you'd create a roughly 1 pico farad capacitor with your body and the 0v terminal.
By VC = q
9 * 1E-12 = 9E^-12 coulombs.
Energy required to charge capacitor ( human body )= C*V^2 = 1E-12*9*9 = 81E-12 J
This is very rough calculation. But the argument is a few pico joules are expended from the battery cell every time the capacitance between the 2 unconnected battery terminals changes.
Is this correct?
To answer this I use the water pressure analogy.
- 2 identical volumes of water.
- different pressures.
-> The one with higher pressure than the other has more water, mass , density.
-> Similarly, the wire at 9V has more charge in it
Me trying to answer :
Everything is a capacitor
The wire is a capacitor
When attached to the 9V terminal, it'll charge up
Also the other terminal is a capacitor, and it's charge will adjust according to the new capacitance of the 9V terminal and connected wire.
Use VC = q
So I reckon yes, the wire is physically and detectably different from its 0V self.
Also, in theory its possible to discharge\ use up a battery just by altering the capacitance of the terminals by connecting random items, e,g, a piece of wire, your finger, a coin, to one terminal but leaving the circuit open.
It would take a long time because the capacitance of such objects with the other terminal is so low hence a low charge hence a low energy usage to charge.
e.g.
I'm guessing by touching the 9V terminal with your finger you'd create a roughly 1 pico farad capacitor with your body and the 0v terminal.
By VC = q
9 * 1E-12 = 9E^-12 coulombs.
Energy required to charge capacitor ( human body )= C*V^2 = 1E-12*9*9 = 81E-12 J
This is very rough calculation. But the argument is a few pico joules are expended from the battery cell every time the capacitance between the 2 unconnected battery terminals changes.
Is this correct?