Electric Fields potential of zero

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Electric potential is not necessarily zero when the electric field strength is zero due to the presence of an arbitrary integration constant in its definition. The electric field is the gradient of the potential, meaning that a constant potential can exist even when the field strength is zero. For example, at the center of a charged spherical shell, the electric field is zero, but the potential is non-zero, indicating that work would still be required to move a charge from that point to infinity. This illustrates that the reference point for potential can be set arbitrarily, affecting the potential values throughout the field. Understanding this concept is crucial for grasping the relationship between electric fields and potentials in physics.
nokia8650
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Can somone please explain why the following is not true:

"Electric potential is zero whenever the electric field strength is zero"

I know that the Field strength is the potential gradient, however why is the above true?

Thanks in advance
 
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Well, the gradient of any constant is always zero isn't it? So why couldn't the Potential be any constant? Certainly zero is one such constant, but there are infinitely many others.
 
Thanks for the reply. Would it be possible for you explain it in a more of a physical context, as I understand that is the case mathematically, however not in a physical context.

Thank you very much
 
nokia8650 said:
Can somone please explain why the following is not true:

"Electric potential is zero whenever the electric field strength is zero"
There are two ways to reason this. One is to realize that there is always an arbitrary integration constant (an example of an electromagnetic gauge choice). The other is to ask yourself, "what is the electric field corresponding to a unifrom potential of, say, 100 V?" and then consider the logical corrolary.
 
Well, to calculate the electrostatic potential is defined only up to some constant. This is due to the definition: \vec{E}=-\vec{\nabla}V. Physically, that means that you can never measure the potential at a point; only the difference in potential between two points can be measured. Typically, when physicists speak of the potental at a point, they mean 'the potential at a point relative to some agreed upon reference point'. Often one takes the reference point to be infinity (i.e. Very very far away from the field point you are interested in) and defines the potential to be zero there. But, you can define the refernence point to be anywhere you like, and the reference potential to be any value you like; so it is true that you can always define the potential to be zero at some point of interest, but that affects the potential everywhere else which is measured relative to that point. So whether or not the potential is zero wherever E=0 depends on where you have set your reference point.
 
nokia8650 said:
Thanks for the reply. Would it be possible for you explain it in a more of a physical context, as I understand that is the case mathematically, however not in a physical context.

Thank you very much
A counter-example to disprove that claim may be given as follows:
Consider the electric field at the centre due to a charged spherical shell. It is 0. What is the electric potential at the centre?
 
Defennder said:
A counter-example to disprove that claim may be given as follows:
Consider the electric field at the centre due to a charged spherical shell. It is 0. What is the electric potential at the centre?

Im not sure how I'd calculate the electrical potential, however there would be some as work would have be done to move a charged particle from there to infinity.

Thank you to all of the above posters for the help.
 
In that case of the charged spherical shell you only need to do a volume integral over the spherical volume for each electric potential contribution due to dV:
So we have dW = \frac{p_v dV}{4\pi \varepsilon_0 r}. Integrate over the volume. A simpler example would be the centre of a charged ring. It also has 0 net electric field strength but non-zero electric potential there.
 
Defennder said:
In that case of the charged spherical shell you only need to do a volume integral over the spherical volume for each electric potential contribution due to dV:
So we have dW = \frac{p_v dV}{4\pi \varepsilon_0 r}. Integrate over the volume. A simpler example would be the centre of a charged ring. It also has 0 net electric field strength but non-zero electric potential there.

Of course, your method implicitly assumes that the reference point has been set to infinity.:wink:
 
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