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"A joule (J) of work is done in moving a coulomb (C) of charge through a potential difference of 1V."

So does 2 volts mean: A joule of work is done in moving two coulombs of charge through a potential difference of 1V?

Or does it mean: two joules of work is done in moving a coulomb of charge through a potential difference of 1V?

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    \$\begingroup\$ So does 2 volts mean: A joule of work is done in moving two coulombs of charge through a potential difference of 1V? ... you said 2 V at beginning and 1 V at end \$\endgroup\$
    – jsotola
    Commented Dec 4 at 18:36
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    \$\begingroup\$ Zuccini - Hi, From seeing your comments on answers, I wonder if the question has now been solved for you? Please see this article about what to do when people answer your question. Although not mandatory, it is encouraged to upvote useful answers (when you have the points to do so) and, if it's solved, to consider áccepting the "best" answer (using your judgement of "best") as that is how to indicate if the question has been solved. (Also see here.) TY \$\endgroup\$
    – SamGibson ♦
    Commented Dec 5 at 13:34

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Neither.

2 joules of work is done in moving 1 coulomb of charge through a potential difference of 2 volts.

More generally:

\$U\$ joules of work is done in moving 1 coulomb of charge through a potential difference of \$U\$ volts.

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Neither.

As C*V=J, the following applies:

1J means 2 coulombs and half a volt. 1J means half a coulomb and 2 volts.

2J means 2 coulombs and 1 volt. 2J means 1 coulomb and 2 volts.

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The equation relating these three units is

joules = volts × coulombs.

So just substitute those numbers in and it is clear which is correct.

xJ = 2V × 1C -> xJ = 2J -> x = 2

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How much of joules of energy a coulomb of charge is experiencing anywhere is that much volts.

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  • \$\begingroup\$ So if the potential difference between two points is, say, 5V then a coulomb of charge at any given point between those two points experiences an energy of 5 joules, right? \$\endgroup\$
    – Zuccini
    Commented Dec 5 at 12:48
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    \$\begingroup\$ @Zuccini The energy is associated with a path, not a point. Moving 1 C of charge between points with a potential difference of 5V either consumes or releases 5 J of energy, depending on direction. \$\endgroup\$
    – John Doty
    Commented Dec 5 at 13:06
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    \$\begingroup\$ @Zuccini 1 coloumb of charge will exprience 1 joules only if there is the potential difference of 1 volt across it. Potential difference is actually the difference in potential. Say we have 2 plates of charge. One plate exerts 10 joules of energy on one coulomb and the second plate exerts 2 joules of energy in one the coulomb in the opposite direction then it means that they have the potential energy of 10-2 = 8 volts. But if the charged plates exerts the energy in the coulomb of charge in same direction then the voltage will be 10+2 = 12 volts. \$\endgroup\$
    – Alex
    Commented Dec 5 at 16:18
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To better understand this, let's think about what voltage actually IS. Voltage comes from volt, which is the unit of Electric Potential. The electric potential of a point in space is the electric potential energy of one coulomb of charge, or how much work it requires to bring one coulomb from some very far away place to that point.

If this is confusing, think of gravitational potential energy, which is likely more familiar to you. It's how much work an object can potentially do based on its height above the ground, i.e. the famous mgh. If you divide that by mass you'd get gravitational potential, or just gh (this is ONLY a function of position and does not depend on whether you put a feather or an elephant there).

When you're dealing with gravity, you arbitrarily assign a point to have 0 potential, based on convenience (the ground, the table, etc). In other words, you don't care about absolute potential, only the difference relative to a fixed point, say the ground. Similarly, you don't care about the absolute Electrical Potential, only the difference between any point in your circuit and the ground. So what voltage is really referring to is the potential difference between a point and the ground.

Now finally onto your question. Going through a voltage of 1 volt is akin to pushing something up one meter. If you have a 1kg object, you do around 10J of work. If you have a 42kg object, you do 420J of work. Going through a potential difference of 2 volts is akin to moving the object 2 meters up. Your same 1KG object will now require 20J of work. It takes 1J to move a 1C charge through a potential difference of 1V, and it takes 2J to move a 1C charge through a potential difference of 2V, or alternately, it takes 1J to move a 1/2 C charge through a potential difference of 2V.

TLDR; Voltage is a measure of electric potential difference, which is similar conceptually to G.P.E. so It takes 2J to move a 1C charge through a potential difference of 2V.

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