Is the Electric Field Always Conservative?

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The electric field is conservative if the line integral is path-independent and equals zero around closed paths. This is demonstrated by the fact that the curl of a conservative vector field is zero, and the electric field can be represented as the gradient of a scalar potential. However, only electrostatic fields are conservative; electric fields induced by changing magnetic fields are not. To experimentally verify if an electric field is conservative, one can use a closed loop of wire with an ammeter; no current indicates a conservative field, while current flow suggests it is not. Understanding these principles is essential for proving the conservativeness of electric fields.
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How can you prove that the electric field is conservative? I've learned about stuff like line integrals but I'm not sure how to prove this particular fact.

Thanks
 
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A conservative vector field has the property that the line integral of the vector field is independent of the path and depends only on the end points. This implies that the line integral of the vector field around a closed path is equal to zero. An equivalent property is that the curl of a conservative vector field is equal to zero. Also, the curl of the gradient of a scalar function is equal to zero. The electric field can be expressed as te gradient of a scalar electric potential.

There are enough different ways hinted at in the above paragraph that I'm sure you can easily prove that the electric field is conservative.
 
Just want to point out that only an electrostatic field is conservative. The electric field induced by a changing magnetic field isn't.
 
Identity said:
How can you prove that the electric field is conservative? I've learned about stuff like line integrals but I'm not sure how to prove this particular fact.

Thanks

If you want to prove experimentally that some electric field is conservative, you can place a closed contour of electric wire, with an in-line ammeter, in the field. If no current flows regardless of the orientation of the wire, then you are dealing with a conservative field. If current does flow, then for selected orientations there is a non-zero emf around the contour and the field is not conservative.
 
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