Voltage vs Electrical Potential Energy Differences? and work

[MrNeuro]

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Im having difficulty understanding the differences between voltage and electrical potential energy

i understand the formulas but i dont get when you can use each formula to calculate the work done

i know that

Δ electrostatic potential energy = qΔV which is work right ?

but i was reading somewhere else that ΔV = Ed which is also equal to work????

which one is right?

and can someone help explain the similarities between voltage and EPE in a gravitational setting...

 

[chiddler]

This is a good question, one that students often get confused about. Electric potential energy (U) can be thought of as energy stored in the the electric field. This energy is just like gravitational potential energy (mgh) and can be converted into something like kinetic energy. The electric potential (V) is similar to the gravitational potential. It is related to the electric potential energy by U = q V, so you can think of the electric potential as the electric potential energy per unit charge.

My TA in an old email.

I think it's U(pot energy) = qV = qΔϕ, where ϕ = electric potential. V = Δϕ, right? Change in electric potential determines the voltage difference. I'm not sure ΔV makes much sense, but correct me if i'm wrong please.

V=Ed is for parallel plate capacitors or an infinitely charged plate that produces uniform Efield. You can't use it for point charges.

 

[MedPR]

Work is qEΔd. Remember that Work=F*Δx, and F=Eq.

Not really sure what else I can say to help though, sorry.

 

[Arctangent]

I generally use W = qΔV to solve for work.
ΔV is in units of Volts or J/C, so the second equation (ΔV = Ed) cannot by itself get you a measure of work (Joules).

As far as comparing E/M work to gravitational work, you have in both equations some fixed point of given charge or mass, distance between the two points (one fixed and one mobile), some displaced point of given charge or mass, and a constant, so the equations are in some sense comparable.

Work is the integral of force with respect to r, so for the gravitational case, you get W = -GM1M2/r, and for the E/M case, you get W = -kq1q2/r, with r evaluated from some start point A and some end point B.

Simplified, for gravitational work, W = mgh via substituting in g, and for E/M work, W = qΔV via substituting in ΔV. So in terms of the second paragraph, where the charge is the quality of a given particle affected by an E/M field, and mass is the quality of a given particle affected by a gravitational field, ΔV can be thought of as the E/M equivalent to (gh). In other words, as (g*h) tells you the per kg change in gravitational potential energy between two points, ΔV tells you the per coulomb change in E/M energy between two points. For both cases, change in energy is work.

Looking at the variables underlying g*h versus ΔV, it may be easier to see that they basically act equivalently. g is related to the mass of the fixed object (usually the Earth), radius of the fixed object (Earth), and the constant G. As for most MCAT problems, we have a fixed radius and mass, g is constant, so in g*h, h or Δr is the only variable. ΔV is based off the charge of the fixed object, the "radius" or distance from the fixed object to the second object at point A, the distance from the fixed object to the second object at point B, and the constant k. As the closest thing the MCAT can get to gravity's "default" of the Earth is maybe a proton or electron's elementary charge, you will often have 2 or 3 variables that can affect ΔV, though as you stated, you may just be given ΔV in the question itself.

So in summary, E/M's Voltage to Electrical potential energy is akin to gravity's (g*h) to Gravitational potential energy.

 

[MedPR]

Where does the V=Ed equation come from?

 

[Arctangent]

V is defined as the integral of E with respect to d, so if we have a region of uniform electric field, then V = Ed. It's kind of like how velocity is defined as the integral of acceleration with respect to time, so under free-fall, a = g, so v = gt.