Recent content by kimchi

just after state of 0 and +Q for the two charges, further flow of charges make them same kind of charge (+charge), repulsion separates them. And no more flow of charge between them happens ><

No OlderDan. You misunderstood me a bit. My view is that when two point charges separated by a distance r attracts each other spontaneously. When they contact, redistribution of charges begins. Electrons flow from Point charge of -Q to Point charge of +2Q, and when it flows -Q to point charge of...

more clear now. thanks my mistake is focus on result=final state. but process is essential too.

-ve induced Q more concentrated at inner surface near the biased + Q in the cavity, any G. surfaces in the shell material enclosed no charges inside, so no flux via them. as elctrostatic equilibium, any conductor is in E=0, so the outer induced +ve Q must evenly distributed to account again...

i guess i know all such thing...but the inner induced -ve charges are not evenly distributed...while outer induced +ve charges are evenly distributed, right? thanks very much

but i ain't sure why should + Q evenly distributed outside the gaussian surface? as what i said above? for ensure the 0 E-field!

+2Q and -Q two pt charges are r apart. AS no external agent exists, they attract to each other and each gains +Q/2 after conduction. Like charges repel. At same distance at r, 2*k(2Q)(-Q)/r = 2*k(Q/2)sq /r + 2 KE... but what i want to know is that when is the like charges repel...just after a...

my interpretation is that...for electrostatic equilibrium the electric field of any pt in a conductor is 0. Inside a G. surface as the outer surface of the shell, electric flux is 0. Image there are no charge inside as induced -ve= +, the induced + charge must be uniformly spreaded outside the...

a metal spherical shell with a +q in it but not centered (in the cavity), why the induced +ve on outer surface is uniformly distributed? (the shell is neutral, and the induce -ve and +ve charges not shown.)