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In 2D why exactly do lines of equal potential exist on a point charge at rest? This doesn't make sense to me if the point charge is hypothetically a lone charge undisturbed by other charges.

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    $\begingroup$ Can you elaborate a bit on what you're asking? Typically we say you can't define the potential at the actual location of a point charge, so we wouldn't draw any equipotential line through the point charge. $\endgroup$ – The Photon Nov 10 at 0:03
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In 3 dimensions, equipotential surfaces are a family of surfaces such that Electrostatic Potential $\phi$ is $\textbf{constant}$ along them. A direct consequence of the fact that $\vec{E}=-\nabla\phi$ is, using this result from Vector Calculus, that the electric field $\vec{E}$ points in some direction $\textbf{normal}$ (or perpendicular) to surfaces $\phi=constant$, i.e., equipotential surfaces. This way, in the case of a point charge, the electric field produced is radial according to Coulomb's Law, so the equipotential surfaces are spherical shells centered at the position of the charge. A physical consequence of the existence of equipotential surfaces is that the Work exerted upon a charge immersed in some electric field is zero along some path contained in those surfaces.

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Equipotential lines are a representation of constant electric potential. Electric potential is given by V = kq/r. So if there was a point charge, then at some fixed distance away from it (r), electric potential will exist.

The reason the equipotential lines for a point charge are propagating circles is that 'r' is the same value at some circle around the point charge, so there will be constant electric potential around the point charge.

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  • $\begingroup$ If that's the case, there is a potentially infinite number of equipotential lines, correct? $\endgroup$ – CalebWilliamsUIC Nov 22 at 6:21

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