# Electric potential VS change in electric potential

JLABBER
Δv=v?
what is the electric potential in a conductor? LETS set the conductor be a shell carrying charge 15nC, then what is the electric potential inside the conductor? should it be just finding v=kq/r? Or we have to find the Δv? Thanks

Homework Helper
Δv=v?
what is the electric potential in a conductor? LETS set the conductor be a shell carrying charge 15nC, then what is the electric potential inside the conductor? should it be just finding v=kq/r? Or we have to find the Δv? Thanks
Electric potential is the energy that a unit charge by virtue of its position relative to some other location. It is always relative.

Electric potential v = kq/r is the work done on a unit charge in moving it from a distance r=∞ from charge q to its position a distance r from charge q. The potential energy of a unit charge at position r relative to a position r', is the difference between the potential energy of unit charge at r relative to r=∞ and the potential energy of a unit charge at r' relative to r=∞.

So in order to answer your question, you have to tell us what you are measuring potential energy per unit charge relative to.

AM

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JLABBER
Well, i have two conditions which both of them have not been so clear for me.

First, two concentric shells (conductors), one with charge 10nC (inside) and another with charge -15nC(outside).Inner shell has radius a and outer shell has radius b.

Then, if we are suppose to find electric field for inside the innerest shell, on inner shell, between the shells, on outer shell and outside both shells.

Which or how are we going to calculate them. We have to start from the outermost shell to find the potential first?
But, for me i do hear that the electric potential is everywhere constant inside the conductors. Therefore, for me ,

the potential outside both shell, i calculate through
ΔV=Vr - V∞=∫-E dr ( upper limit be r and lower limit be ∞)=k(q1-q2)/r-0=k(10-15)n/r.

the potential on outer shell, Vb=k(-5)n/r=k(-5)n/b.
the potential between the shell, ΔV=Vr-Vb=∫-E dr ( upper limit be r and lower limit be b)=kq1/r-kq1/b. Thus, rearrange, Vr=kq1/r-kq1/b+Vb=k10n/r-k10n/b+K(-5)n/b((This part i start to be blur))

Then the potential on shell a, Va=k10n/r-k10n/b+K(-5)n/b=k10n/a-k10n/b+K(-5)n/b

Next, then i have little idea of how to continue...
(this part start to knock here and there)

Lets see this example first before proceed to the next condition. Thanks for your concern.

(Additional out of topic question: when are you online most frequently. May be we can have a chat to clear the blurry faster)

ZealScience
Since electric potential is the electric potential energy per unit charge, it is relative. Potential energy is always relative.

Your equation V=kq/r, to my perspective, is relative to point of infinity where potential is zero, since V=∫E·dx(from infinity to r)=∫kq/x^2·dx=-kq/x|(r, ∞)=kq/r. But I don't think that calculating potential from infinity is not handy in calculation of circuit question.

JLABBER
yes, i know that. That is why i feel the next following solution become not so correct.

the potential on outer shell, Vb=k(-5)n/r=k(-5)n/b.
the potential between the shell, ΔV=Vr-Vb=∫-E dr ( upper limit be r and lower limit be b)=kq1/r-kq1/b. Thus, rearrange, Vr=kq1/r-kq1/b+Vb=k10n/r-k10n/b+K(-5)n/b((This part i start to be blur))

Then the potential on shell a, Va=k10n/r-k10n/b+K(-5)n/b=k10n/a-k10n/b+K(-5)n/b

Next, then i have little idea of how to continue...
(this part start to knock here and there)