Hi again raja,
Lemme try to elaborate a bit more...Say you have two ideal voltage sources whose -ve terminals are connected to GND & +ve terminals are hanging as floating(Not connected as of now, No connections, no leakages, no arching nothin...). Then you can assume these points to be ,say, A = 2V + terminal , B = 3V +ve terminal) and assert that these act as POINT CHARGES and have separate electric fields of their own...say k.q1/r
2 and k.q2/r
2 But force exerted on one due to another will always be same --> k.q1.q2/r
2...Since the reference GND wrt which the Potentials of points A and B are different [recollect V = k.q/r], this itself means that the amount of charge present at these nodes is different...So I guess that explains the different charge densities at both the points..
Now, as soon as you connect them through a conductor, The electric field between the two points becomes the -ve gradient of the potential between two points. [It always is the -ve gradient of pot diff b/w 2 points but till the time they were unconnected, we could have argued that individual E fields exist and are differentiable]. Now the E field becomes (3V-2V)/ Length of the conducting wire...
But, what if the connecting wire is a superconducter ? ZERO resistance...Perfect e- gas....
Two processes will come into play -->
1> Diffusion process ---> as i explained earlier...Charge carriers e-s diffuse from higher density to lower density -- until dynamic equilibrium is attained...without any external force...
2> E Field b/w two points ---> As soon as you connect the nodes, There will exist a E field between two points which will point from +ve node to -ve node (3V node to 2V node) because of different charge densities at both the points...[remember from prev post..why did E field generate in d 1st place ?? --> coz of diff charge densities.] This E field will cause e-s to move from 3V end to 2V end...reducing E field magnitude in the process..until finally an equilibrium is attained when both potentials become equal...
Both there processes will ensure that the superconducting wire...will essentially settle at a common voltage...(ultimately making both the points appear as SHORTED)...But what makes them a SHORT ?? --> These two processes....
If the wire has a finite resistance...Then also initally there will be a voltage drop across the wire...But eventually, both the ends will settle to same voltage levels, because of the above mentioned processes...Differnece being that In this case, the diffusion process will be confronted by the finite resistance of the wire...
Hope that explains it all....Still welcoming suggestions....These are ultra-basic topics involving physics...any1 can be wrong anytime :P
Anyways, Did this help raja or made it even more confusing
?
cheers,
Mayank....