Electrostatic Potential Energy with Constant Electric Field vs NonConstant Field

[Hereformore]

When determining potential energy we have the relationship:

PE = V*q = E*q*d

But depending on the scenario we can't use them interchangeable right?

If you want to calculate the work needed to be done by a particle going from a distance Y to right next to another particle of the same charge, then you couldn't use PE = E*q*d since the electric field wouldn't be the same as the particle got closer and closer.

But you could use the voltage relationship here if you calculated the voltage difference between the two points.

In what situation would you be able to use the PE= E*q*d? Where there is a constant electric field between two particles? (so not between a parallel plate capacitor).

 

[zoki85]

In what situation would you be able to use the PE= E*q*d? Where there is a constant electric field between two particles? (so not between a parallel plate capacitor).

In situations where constant external field acts on a particle

 

[paulina]

Hereformore,
don't u think that with the change in the electric field, there is also a change in the distance... so the net result of E*d*q can be constant. So why do u need to have a constant electric field to use the equation ?

 

[GergM]

If you want to calculate the work needed to be done by a particle going from a distance Y to right next to another particle of the same charge, then you couldn't use PE = E*q*d since the electric field wouldn't be the same as the particle got closer and closer.

You cannot calculate the work of the electric force which cause the movement of a charge between two points with PE = E*q*d, not because the not uniform field, but because the work you are looking for, is the work between two points, and the PE is the work between a point and infinit.(were we set for convenience zero (Nullpunkt), and every value of potential is measured with respect to that zero)

 

[Khashishi]

Basically, your equations are the same, except that V is replaced by E*d.
That's because V = E*d, (or more conventionally, V = -E*d) for cases in which E is constant. Actually, a parallel plate capacitor is a good example of where you can use this equation, since a parallel plate capacitor has a roughly constant E field between the plates, as long as you stay away from the edges.

This should all become very clear with calculus-based physics, where V = -E*d is just a special case of
$V = -\int \mathbf{E} \cdot \mathrm{d}\mathbf{s}$
which is valid when E is varying along the path.