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!

Grounding location of a metal object

  1. Sep 19, 2010 #1
    if I have 2 spheres A and B initially neutral, isolated from ground and in contact with each other, when I bring a negatively charged rod near to sphere A, I understand that positive charges will be induced on A and negative charges on B. Suppose I were to ground the sphere (a wire either touching A or B), does the location of my grounding affect the final distribution of charges?
     
  2. jcsd
  3. Sep 19, 2010 #2

    EHT

    User Avatar

    Since the charges tend to form a configuration with the smallest potential energy, the final configuration doesn't depends on the location of grounding. Any potential hill can always be climbed, due to thermal fluctuations of the electrons

    sorry for my bad english
     
  4. Sep 19, 2010 #3

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    It does. With the spheres ungrounded and isolated, the charges will separate with the opposite charges attracted towards the charged rod and the like charges moving away. The result is as you describe with the charges bunching up on opposite surfaces. A ground is a charge sink/source and it acts like a conductor. Thus, the charges can move freely into and out of the ground in response to an induced voltage. This means that the like charges that moved to the opposite surface can now be repelled into the ground. The result is that now the spheres (assuming they remain in contact) now have a net charge (because you have driven the like charges off of the spheres down the ground connection).

    Let's just take a single sphere because we do not need to think about them moving about. If I have a neutral sphere and bring into its vicinity a negatively charged rod, then the positive charges collect on the sphere's surface nearest to the rod and the negative on the opposite surface. Then I ground the sphere and this allows the negative charges to be driven down into the ground (the ground is one big conductor so there is no work needed to move charges into the ground). Now if I removed the ground and moved the rod away without allowing it to make contact then the sphere will be posivitely charged.

    Another way to think of it is that the ground always pulls a conductor to 0 V. So instead of a sphere and rod, we have a parallel plate capacitor. One plate is charged up to +1 V with respect to the ground and brought near our second plate. Then we ground the second plate so that it has 0 V to ground. Thus, now the plates have a potential difference of 1 V and the +1 V plate has positive charges and the 0 V plate has negative charges. But since the ground is an infinite source/sink, it is considered to be always charge neutral no matter how much charge we drive into or out of it. So the 0 V plate and ground together are neutrally charged. But if we remove the ground from the 0 V plate while keeping the 1 V plate in place, the negative charges on the previously 0 V plate are still there and thus the plate is now negatively charged. So this is another way to see how the end result is that the conductor becomes charged.
     
  5. Sep 19, 2010 #4

    EHT

    User Avatar

    @Born2bwire: for the induct-tor, you are using negatively charged rod in the first case and positively charged conductor in the second case. Which is compltely different, if we use the same induct-tor it doesn't matter where do we put the grounding cable on the sphere.
    The negatively charged rod will induce positive charge on the sphere if grounded and The positively charged rod will induce negative charge on the sphere if grounded
    regardless of where we touch the sphere using the grounding cable
     
  6. Sep 19, 2010 #5
    @Born2beWire: It does. With the spheres ungrounded and isolated, the charges will separate with the opposite charges attracted towards the charged rod and the like charges moving away. The result is as you describe with the charges bunching up on opposite surfaces. A ground is a charge sink/source and it acts like a conductor. Thus, the charges can move freely into and out of the ground in response to an induced voltage. This means that the like charges that moved to the opposite surface can now be repelled into the ground. The result is that now the spheres (assuming they remain in contact) now have a net charge (because you have driven the like charges off of the spheres down the ground connection)....

    Ah yes but my question (as EHT has kindly replied) was whether there would be a difference where I put my grounding wire. If I use the negatively-charged rod to the induce the charges on the spheres (positive->nearer and negative->further), it seems unintuitive that the same effect would occur (i.e. the dissipation of the negative charges on the further sphere to ground) even if i put the ground wire right on the surface where the positive charges are(nearer sphere). But I guess since your reply makes no mention of where you place the grounding wire that you also think that location is irrelevant?
     
  7. Sep 22, 2010 #6

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    It does matter a bit but not in the way you are thinking. The ground forces the entire conductor to be in the same potential regardless of the contact point so we should expect that the ground point is immaterial. However, the grounding rod itself interacts since it changes the geometry of the structure. The way that we place the grounding rod does affect the amount of charge that becomes induced due to the fact that it can contribute to the capacitance of the situation.

    For example, we have our parallel plates again and make one of them smaller in area than the other. If we place our rod parallel to the surface of the small plate and touch the side then we effectively increase the surface area of the plate and increase the capacitance of the entire system. If we just place the rod perpendicular on the back side then the change in capacitance is very minor. Still, the effects should be rather minor and probably negligible in a first order approximation. That is, for the parallel plate problem, the charge distribution over the plate will not change unless we account for the field fringing effects.

    But as long as we ignore these minor effects, we should just realize that the grounding of the objects remains the same. Just assume that we use a very thin wire to connect the ground and that the physical presence of the wire will be very minor.

    So the way to look at it here then is that if we place the ground point on the side close to the inducing object, then what we do is we draw up opposite charges from the ground. That is, place a negatively charged rod near the sphere. If we place the ground on the opposite side of the charged rod then we drive negative charges from the sphere down to the ground. But if we place the ground on the same side as the inducting rod, then we draw up positive charges from the ground on to the sphere since they are being attracted by the charged rod. Either way, we end up leaving a net positive charge on the sphere when we remove the ground.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook