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I read that a negative charge that gains potential in a field loses energy I know the formula that relates voltage and charge to joules. But I can't visualize it. Where are the equipotential lines? Are we to assume that the field that the charge is in is defined by a positive source charge? If so, then why should the test charge gain potential by getting closer to the source charge? Graphs show equipotential increasing as the source charge is approached, no? But certainly as the negative charge approaches a positive one, the potential should decrease because opposites attract. Once again, I have a hard time visualizing the equiposition lines: What does a negative charge look like in a field where it is gaining voltage in that field? Pictures would be nice.

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  • $\begingroup$ You are falling into a terminological confusion between the "potential energy" and the "electrostatic potential." $\endgroup$
    – Buzz
    Dec 22, 2020 at 21:15

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why should the test charge gain potential by getting closer to the source charge?

A negative charge has higher potential energy when it is at a lower potential and vice versa.

Why? Because its charge is negative.

This allows us to define one electrostatic potential function independent of what test particles might later be introduced to the system and whether their charges or negative or positive, instead of having to have a "electrostatic potential for negative test charges" and a different "electrostatic potential for positive test charges".

But certainly as the negative charge approaches a positive one, the potential should decrease because opposites attract.

Yes, as the negative charge approaches a positive fixed charge it moves to a position of higher electrostatic potential, and therefore it (or rather, the system) loses potential energy.

Where are the equipotential lines?

That depends entirely on what charge configuration is producing the potential field.

Regardless of what they are, whether you move a positive charge along the equipotential lines or a negative charge, the system's electrostatic potential energy won't change.

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