Question

Please look at my answer and make corrections.

1. It is not possible to have an electric field line be a closed loop. Explain why this is true using potential and/or equipotential lines.

Hints:

1. The focus should not be on talking about charge distributions.

My answer:

Because the force of electric fields is conservative, the change in potential energy from arbitrary points A to B should be the same magnitude as B to A regardless of the path taken. A closed loop for an electric field line implies that if a charge moved through the field line, it would begin and end at the same location and would thus have a different potential energy because it followed the line. This would be a result of the fact that in this imaginary closed loop, the displacement would always be parallel to the field (as opposed to perpendicular like we want it to be in relation to the equipotential lines). Because of this, the work done will always be positive and, because by this definition it can’t ever be 0, the fact that the force is conservative is false (which we don’t want it to be).

Answer 1

Below is the answer based on equipotential surfaces.
From the concept of Equipotential Suw.face, which is perpendicular to the electric field. from hypothetical point of view if eclechic field line be a closed loop. Circumference of eisele showing field lines. 4 AB, CDi EF, GH ore perpendicular to electic field. then AB (Electic field arrection at A & B is differenct So A & B are not in same potental. In AB at o Electic field zero so, all the point AIB and a

has different potential. likewise coiffi GH will also have different polen Hals in their Une. So, if field tines are closed loop it is not possible to get equipotential Susfaces. If somehow assume AB,CO, EF, GH is equipoten tal surface we know equiporenlial surfaces mever intereseet each other and always pasellele here AB, CRIEF, GH intersect at a. which violates the equipotential conditons. Hence, elechic #the field line will not form closed loop.