Voltage Confusion: What Am I Missing?

[Jaccobtw]

Homework Statement

For two oppositely charged particles, does the distance between them increase or decrease voltage?

Relevant Equations

V = kq/r

I'm confused about voltage. According to this equation, decreasing distance would increase the voltage between two oppositely charge particles, but doesn't increasing the distance increase the potetnial energy between the two particles because the amount of work done would increase with distance, correct? It takes more energy to separate a charge a greater amount of distance increasing the voltage. Also the particles would have greater kinetic energy beginning from a larger distance.. What am I not understanding here? Thank you.

 

[BvU]

Hi,

I'm confused about voltage.

Perhaps it helps to consider the electric potential field as energy per unit of charge.

According to this equation, decreasing distance would increase the voltage between two oppositely charge particles,

And not talk of voltage between particles.

So a single charge $q_1$, located at the origin of a coordinate system, has an electric field potential associated in space according to $V = {k\,q_1\over r}$.

To bring a test charge $q_2$ (a second charge) from infinity to a position at distance $r$ from the origin requires an energy of $q_2V = {k\,q_1q_2\over r}$.
I.e. work has to be done to do this if the charges are of opposite sign.

[edit] OOPS! Someone should have jumped on this !
I.e. work has to be done to do this if the charges are of the same sign.

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