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Potential of a Grounded Conductor 
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#1
May908, 02:56 PM

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Hi,
As I understand it, the potential of a charged conductor is the work done in bringing a unit positive charge to its surface from infinity. Is this correct? When a conductor is grounded, then any excess electrons on it go into the ground. Is this correct? If so, why? Assuming that they do go into the ground, the conductor now has no charge and therefore produces no field. So the integral of E dr from infinity must be zero. Is this why the potential of a grounded conductor is zero? 


#2
May1008, 02:58 AM

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That is correct.
The “ground” or the earth has a huge capacitance, and can absorb any charge, positive or negative, in the order of magnitudes that we deal with. It can be considered as an infinite sink or a reservoir. Its potential can be considered to be zero for all practical purposes. (If I remember rightly, for grounding, especially in high tension cables, the neutral wire has to go deep into the ground and generally touches the water table, wherever possible, for fast dissipation of charge.) When a conductor is grounded, its potential becomes same as that of the ground, i.e., zero. If the conductor is not in any external electric field, then the excess charge flows to the ground to make the conductor’s potential zero, and that is why the integral will be zero, if the potential is set to be zero at infinity. Suppose some positive charge is induced on one side of it, and so there is negative induced charge on the other side. Now if it is grounded, the free negative charge flows to the ground but the induced positive charge remains. The potential of the conductor is still zero, because the potential of the conductor is now the sum due to the external field and the induced charge. So, excess charge need not always flow to the ground, but the potential of a grounded conductor is always zero. (That's nice! Forgot that the library has been opened. It's linking the phrase "Electric Field", but does it have to make it title case always?) 


#3
May1008, 03:53 AM

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#4
May1008, 05:39 AM

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Potential of a Grounded Conductor
If there is only one charge under consideration far away from other charges, then at any point the potential is due to that charge only. If there are two charges, then due to the principle of superposition, the potential at a point is sum of the potential due to both the charges. We may consider the potential due to one charge separately, but the actual potential is due to both charges.
If there is a continuous distribution of charges, then the actual potential and field at a point will be due to the whole distribution, and we have to sum or integrate at a point to find the total potential or field. If a field is given in a situation, that is, we are not bothered with the source of the field, and some charges are there, then the field at a point will be the given field plus the field due to the charges. This is always the case, and conductors are no exception. Perhaps you are not so familiar with problems dealing with continuous distributions of charge, and that is why this confusion. 


#5
May1008, 05:51 AM

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I understand that the actualy potential is the integral of the actual field (i.e. the total field due to all charges). But when we say "the potential of charge one", we are only talking about the integral of its field, right? Then when we have more than two charges, we say that the potential is the sum of the potentials due to each charge. In the case of conductors, when we say "the potential of the conductor", you said that we're also considering the external field. So there seems to be a difference in what we mean by "the potential of A", depending on what A is.



#6
May1008, 09:41 AM

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A conductor has the same potential at every point once the charges come to an equilibrium. So, a grounded conductor has potential zero throughout (provided we have chosen the potential thus). This is why we can use the phrase potential of a conductor. That is the difference between talking of the potential of a charge and potential of a conductor.



#7
May1008, 12:44 PM

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Except for superconductors, current carrying conductors contain a voltage gradient. Ground is a reference potential. Connecting The + or  terminal of an open circuit battery to ground results in no electrical energy change.



#8
May1008, 02:36 PM

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(I wonder if people really read the whole thread...)
We were talking about electrostatics here, as was quite evident from the discussion. Also, since ground is just a reference potential, it's convenient to set it to zero. What does that mean? Has the ground become of potential V or is the terminal at zero potential now? 


#9
May1008, 11:37 PM

P: 73

Unfortunately, voltage is called potential. Potential has the implication of potential energy. A single contact point on a voltage gradient contains no voltage's energy, because storage of voltage's energy requires a voltage gradient. In other words, an open circuit wire connected to a battery terminal does not acquire point charges from the battery.
Within a closed circuit, the wire can contain energy in the form of magnetism and voltage. 


#10
May1108, 03:27 AM

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OK, we'll use use the term potential from now, and it will mean what it means in Physics. Try to answer the last question of my last post. I don't think there is there is semantics involved there.



#11
May1108, 07:42 AM

P: 73

OK, I'll try. For a point charge, the energy of bringing a like point charge to a particular distance from the charge is potential (energy). An OBJECT that contains a voltage gradient has a particular number of point charges; current does not change that number (Kirchoff's law).
Within current, an external electron deposits energy w/o remaining with the object. In other words, adding potential energy is adding only the energy of an external electron. Adding an external electron to an object does not necessarily increase voltage. Note: The definition of voltage as energy per charge corresponds to acceleration energy (ELECTRONVOLTS) per charge (U = QV). Voltage cannot be defined as charging energy per charge in a capacitor (U = 1/2 QV). 


#12
May1108, 09:31 AM

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I am not able to understand whether this is an answer to my question. I state my question again below.
When a current is not flowing through a conductor at any point, then the potential of the conductor must be constant at all points. You have joined the earth and the positive terminal of an open battery by a conducting wire. You have said that current will not flow through this wire, which is correct. If before joining, the terminal of the battery had a potential of V with respect to the ground, then what is the potential now of the battery terminal, the wire and the earth, in terms of V? (No explanation needed.) 


#13
May1108, 02:05 PM

P: 73

I'll try again. First I'll explain current.
Let us use only superconductive wires. Wires join capacitors together when a closed circuit exists. Resistors can contain voltage, voltage has voltage's energy, thus resistors are capacitors. Upon creations of an open circuit, the voltage in a resistor drops to zero. If current (instead of voltage) converts to thermal energy, an open circuit would not cause resistor volts to drop to zero. Upon creation of a closed circuit, source voltage appears across the superconductor. The superconductor is an inductor, all capacitors (including resistors) short their voltage to the inductor. Afterwards, the magnetic energy in the inductor converts to voltage's energy in each of the series capacitors, with inductor voltage dropping to zero. The voltage that was across the inductor becomes distributed among the other series components, keeping the voltage to resistance ratio (current) in each component constant. Now I'll answer your question. When a wire connects a battery terminal to (conductive) ground, the ground can be a series component of a closed circuit. 


#14
May1108, 02:44 PM

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If before joining, the terminal of the battery had a potential of V with respect to the ground, then what is the potential now of the battery terminal, the wire and the earth, as a function of V? (No explanation needed.) (Think a while; I don't think you need superconductors. ) 


#15
May1108, 10:38 PM

P: 73

By definition ground voltage is zero. The battery terminal that is not connected to ground has voltage V. If the terminal with voltage V is also connected to ground, then the terminals are shorted together.
Even if voltage's energy in a battery shorted with a superconductive wire, voltage's energy will eventually approach zero as voltage's energy converts to other energy. I suppose the terminal that initially had V volts, could have any amount of volts between zero and V after shorting, depending on elapsed time and characteristics of circuit components. 


#16
May1208, 03:46 AM

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#17
May1408, 03:39 PM

P: 73

When the + terminal of a battery is grounded, that terminal has zero volts (by definition). Afterwards, the terminal then has V volts.
Point charge position does not determine voltage's energy. In a Wimshurst generator, relative movement of voltage gradients cause voltage's energy to continuously increase. This continuous energy increase occurs within an open circuit (except during sparks) and is beyond the energy that can be due to force between the initial sets of point charges. Note: Each pair of Wimshurst generator plates is not a capacitor. Without neutralizer bars, sparks occur each half disk turn, regardless of number of plates on each disk. The voltage gradient's position within each disk remains stationary within the disk. With two neutralizer bars, the voltage gradient remains diagonal as the disk spins. Magnetism redistributes voltage's field each time a pair of the disk's plates form a closed circuit with a neutralizer bar. 


#18
May1808, 07:38 PM

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(Sorry for the delayed response.)
In this situation, the negative terminal will have a potential of (–V), if V was the emf of the open cell. 


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