1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Electric Potential reference value (or zero).

  1. Feb 3, 2017 #1
    Hello forum members,

    The electric potential for a point charge is a scalar function given by $$V = \frac {kq}{r}$$

    This means that the potential has a nonzero value everywhere. The potential becomes ##V=0## when ##r=\infty##. However we know that what matters is the potential difference ##\Delta V## and not the absolute value of potential at each spatial point. This is because the physically important and measurable quantities like force and electric field depend on that difference and not on the absolute value of V at each spatial point...

    How would we make the potential to be ##V=0## not at infinity but at different spatial point, like ##r= 5##? How do we modify the function ##V = \frac {kq}{r}##? Like this
    $$V = \frac {kq}{r-5}$$

    Is that it?

    Thanks,
    fog37
     
  2. jcsd
  3. Feb 3, 2017 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, that changes the functional behaviour of the potential, it would diverge at r = 5 instead of r = 0. What is physically irrelevant is a constant addition to the potential.
     
  4. Feb 3, 2017 #3
    Oh, you are right.

    For instance, something like this: $$V = \frac {kq}{r} + 5$$

    So at ##r=\infty## the potential is 5 and wherever the term ##V = \frac {kq}{r}## equals ##-5## the potential will become ##V=0##

    Thanks
     
  5. Feb 3, 2017 #4

    jtbell

    User Avatar

    Staff: Mentor

    The gravitational potential is similar. With the reference point at infinity, the gravitational potential outside the Earth's surface is ##V = -GM/r## where ##M## is the mass of the Earth and ##r## is the distance from the center of the Earth. You can reset the reference point to make the potential zero at the Earth's surface by adding a constant: $$V = -\frac {GM} r + \frac {GM} R = GM \left( \frac 1 R - \frac 1 r \right)$$ where ##R## is the radius of the Earth.

    Optional exercise: let ##r = R + h## where ##h## is the height above the Earth's sufrace, and show that if ##h \ll R##, then ##V \approx gh## (note little ##g## not big ##G##), so the potential energy of a mass ##m## at height ##h## is ##\approx mgh##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted