How do I apply Maxwell's equations?

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Lasha
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For example, if I have a magnetic field perpendicular to some surface and I change this magnetic field with constant speed, how do I calculate the Electric field at any point on this surface, since ∫E⋅ds=k, where k is some constant, could be done with many different vector fields.
 
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Or does E always equal k/s cause ∇⋅E=0 where there's no charge?
 
Are you talking about a mathematical surface or a physical surface? A physical material can have charge and may be made of dielectric,diamagnetic materials. For some cases, you need to apply some boundary conditions and then solve Laplace's equations. But for simple cases, you can often look at the symmetry of the problem and apply Gauss's law.
 
Khashishi said:
Are you talking about a mathematical surface or a physical surface? A physical material can have charge and may be made of dielectric,diamagnetic materials. For some cases, you need to apply some boundary conditions and then solve Laplace's equations. But for simple cases, you can often look at the symmetry of the problem and apply Gauss's law.
I don't get it how do I solve it with Gauss or Laplace when ∇×E≠0. I don't have a charge or even a region where electric field is made by a charge.I simply have sum of many circular vectors of E at any point on this surface.
 
There are many possible fields because the problem is under-specified. You could have a Helmholtz coil which creates a varying magnetic field in a space, and you could have charges outside the region of interest. The charges would produce a static E field which overlapped the E field due to the Helmholtz coil. Since this is a valid physical situation, it's a solution to Maxwell's equations. So you can have all sorts of different static E fields superimposed with of the Helmholtz coil field which are all solutions to Maxwell's equations. This is why you need to specify boundary conditions.

If you are looking for the solution which is just due to the Helmholtz coil itself (no static fields from charges outside the region of interest), then you can apply symmetry to Faraday's law to get E.