Why do we have a charge in the denominator of equation for voltage?

In summary, the charge in the denominator of equations for voltage and electrostatic potential is not dependent on charge. It was originally included in the derivation of the formula for voltage, but it is not necessary and can be simplified by using a scalar field. This concept is similar to gravitational potential, where the mass is defined as the work needed to accelerate an object to a given speed.
  • #1
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Homework Statement
Why do we have a charge in the denominator of equations for voltage and el. potential if both voltage and el. potential are not dependent on charge?
Relevant Equations
U=W/q, fi=Eep/q (fi=el. potential, Eep= el. pot. energy, U= voltage)
Why do we have a charge in the denominator of equations for voltage and el. potential if both voltage and el. potential are not dependent on charge?
Is it just because that was the only way to derive the formula for voltage and then we realized we don't need q? U=W/q --> U=eqd/q.
 
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  • #2
It's a definition. Electrostatic potential ##V_e## is electrostatic potential energy ##U_e## per unit charge. The energy does depend on the charge but it is easier to think of a scalar field ##V_e## such that when we place charge ##q## at some point in space, its energy will be ##U_e=qV_e##.

You have already encountered this idea. Compare with something familiar, gravitational potential. Near the surface of the Earth it is ##V_g=gh##. When one puts mass ##m## at height ##h##, its gravitational potential energy is ##U_g=mV_g=mgh.##
 
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  • #3
To add to @kuruman's reply…
It depends what you take as fundamental. If you take energy, distance and time as fundamental then you would define the mass of an object as the work needed to accelerate it to a given speed.
 

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