2 equations for electric field

In summary, the electric field can be described by both -∇V and -∂A/∂t equations. Gauss's law is used for electrostatics and Faraday's law for magnetism. In electromagnetism, the Maxwell equations are used which state that the electric field must satisfy all equations at all times. The magnetic field can also be defined by the curl of A, which follows from another Maxwell equation stating that the divergence of B is always zero. This is always satisfied, but there is a lot of discussion about it as it implies the absence of magnetic monopoles. The equation for the curl of B includes both the current density J and the change in E with time. In the case of time-independent
  • #1
Isaac0427
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Hi guys!
I have noticed that the electric field has 2 equations to it- the negative gradient of the potential, and the negative partial derivative of A with respect to time. Is this like e=mc^2 and e=hν, where only one equation can be used at a time? Can an electric field be described by -∇v OR -∂A/∂t (in which case you are getting an electric field from changing a magnetic field).
 
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  • #2
Good observation !

At school we use Gauss's law to find E (or the potential V) for given charge distributions (that's usually the electrostatics chapters) and use ##\vec E = -\nabla V## to link the two.

And in the magnetism chapters we encounter Faraday's law of induction to find an e.m.f. (again, a V and an E!) as a consequence of a changing magnetic field.
So yes, the corny answer is: yes the field can be described by both.

Further on in the curriculum we have electromagnetism and things are heavily intertwined and we have the Maxwell equations

But you are right: the electric field has to satisfy all the Maxwell equations at all times.
 
  • #3
If it is an electrostatic field, only then it can be equal to the negative gradient of a potential. If it is changing with time, then you need a term -∂A/∂t to calculate the electric field. A is called the vector potential, which is also related to the changing magnetic field that you talked about.
 
  • #4
BvU said:
Good observation !

At school we use Gauss's law to find E (or the potential V) for given charge distributions (that's usually the electrostatics chapters) and use E⃗ =−∇V\vec E = -\nabla V to link the two.

And in the magnetism chapters we encounter Faraday's law of induction to find an e.m.f. (again, a V and an E!) as a consequence of a changing magnetic field.
So yes, the corny answer is: yes the field can be described by both.

Further on in the curriculum we have electromagnetism and things are heavily intertwined and we have the Maxwell equations

But you are right: the electric field has to satisfy all the Maxwell equations at all times.
Is there another way to define the magnetic field other than the curl of A. I know from Ampere's law, the curl of B relies on two things: the change in E with time, and the current density J. Is it like the electric field, where only one of those can describe any given magnetic field. Would that mean there is another part to B=∇XA? I may be completely off, but it seems as though there are 2 mutually exclusive ways to define each of the fields in electromagnetism. I'm sorry if this is a stupid observation.
 
  • #5
There are two "equations" here. The dependence of the curl of B on dE/dt and the current density J, is one of Maxwell's equations and is always satisfied. This has nothing to do with writing B as the curl of A. The relation B = curl A follows from another Maxwell equation, which says that the divergence of B is always zero. This is also always satisfied, though there is a lot of discussion about it, because that Maxwell equation says that there are no magnetic monopoles. If we accept that, then the divergence of B is zero, and B can always be written as the curl of a vector field, which is called the vector potential A. This part has nothing to do with time dependence. Whether B is time dependent or not, you can always write it as the curl of A.
 
  • #6
Chandra Prayaga said:
There are two "equations" here. The dependence of the curl of B on dE/dt and the current density J, is one of Maxwell's equations and is always satisfied. This has nothing to do with writing B as the curl of A. The relation B = curl A follows from another Maxwell equation, which says that the divergence of B is always zero. This is also always satisfied, though there is a lot of discussion about it, because that Maxwell equation says that there are no magnetic monopoles. If we accept that, then the divergence of B is zero, and B can always be written as the curl of a vector field, which is called the vector potential A. This part has nothing to do with time dependence. Whether B is time dependent or not, you can always write it as the curl of A.
But is ∇X(∇XA) equal to J OR ∂E/∂t or J PLUS ∂E/∂t?
 
  • #7
The plus sign. See the Maxwell equations link.
 
  • #8
BvU said:
The plus sign. See the Maxwell equations link.
Correct. Maxwell equations are always correct, so curl (curl A) = J + ∂E/∂t. Particular case of that equation: If the fields are time independent, then the second term is zero and curl (curl A) = curl B = J, which is Ampere's law.
 

1. What are the two equations for electric field?

The first equation is the Coulomb's Law, which states that the magnitude of the electric field between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The second equation is Gauss's Law, which relates the electric field to the charge enclosed within a closed surface.

2. How do you use Coulomb's Law to calculate the electric field?

To use Coulomb's Law, you need to know the magnitude of the two charges and the distance between them. Then, you can plug these values into the equation: E = k * (q1 * q2) / r2, where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

3. What is the significance of the inverse square relationship in Coulomb's Law?

The inverse square relationship in Coulomb's Law means that the electric field strength decreases as the distance between two charges increases. This is because the electric field lines spread out in all directions as they move further away from the charges, resulting in a weaker field.

4. How do you use Gauss's Law to calculate the electric field?

Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. This can be written as ΦE = qenc / ε0. To calculate the electric field, you can rearrange this equation to solve for E: E = ΦE * ε0 / qenc.

5. Can these equations be used for any type of charge distribution?

Yes, both Coulomb's Law and Gauss's Law can be applied to any type of charge distribution, whether it is a point charge, a line charge, or a surface charge. However, the calculations may become more complex for non-point charges, as they require integration to determine the total charge enclosed or the electric flux through a surface.

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