# Voltage Formula (Potential Diff) [closed]

1 volt = 1 joule / 1 coulomb

#1 Why in this equation is there a need for volt to be divided by charge ? Because according to me voltage is the potential difference between 2 points in a circuit , and it is the energy to push electrons from the surface of the conductor(wire) to start flowing inducing current . it's simply the work done or the energy , Why divide by energy ?

#2 Could anyone derive the formula for voltage?(Potential Diff)

#3 In resistive circuits why is voltage proportional to ampere because voltage = work done/ charge so inorder for the voltage to increase , the work done has to be increased or the charge has to decreased and if charge decreases the current decreases ( I = Q/t ) ..

Could someone help me solve these basic doubts and fill the gaps in my understanding ?

• Please try to focus on one question at a time and don't ask multiple at once.
– noah
Commented Mar 2, 2022 at 12:18
• These are all very basic electrical questions. Have you taken any courses? Commented Mar 2, 2022 at 13:02
• Nope , just high school classes Commented Mar 3, 2022 at 4:20

Voltage isn't the energy needed. Voltage is defined as the work needed to move a unit charge from some reference to r ,against the electric field.

The term "unit charge" isn't specific to a specific charge, it is the work done per unit charge (aka \q)

With this definition set, how can one derive a formula for this?

Well from the very definition of work,

The amount of work done BY the field on charge q is just

$$\int \vec{F} \cdot \vec{dr}$$

The work done PER UNIT CHARGE is this expression divided by q

$$\int \vec{E} \cdot \vec{dr}$$

If this represents the work done BY the field, then what is the amount of work that would have to be done against the field? Well it would be the negative of this expression, as if the field does ="-w" amount of work per unit charge in moving from ref to r, I would have to therefore do "w" amount of work against the field, to move it there.

So we have

V = -$$\int_{ref}^{r} \vec{E} \cdot \vec{dr}$$

What is "ref" Well traditionally it is chosen to be infinity, however this is irrelevant as all values for "ref" are equally valid and all give the same change in potential as they differ by a constant (c-c=0)