# Is electrostatic potential energy of a charge is different depending on whether the electric field attracts or repel the given charge?

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Yes and No.

Yes they are different in the mathematical sense. No because the magnitude would be the same.

For an example, take a look at something that is familiar:

Gravitiational Potential Energy. To lift a mass at a constant rate, the amount of work done on the system by an external force (provided by a person lifting, for example) is how the Gravitational potential energy changes.

This can be stated as a form of the first law of thermodynamics

W = ∆PE

Where W > 0 when you work *against* the field, and W<0 when work *with* the field. - Say we release the mass, the field causes the mass to accelerate *with* the field.

mass reacts to a gravitational field much like a *positive* charge interacts with an electric field. (Remember, field lines are defined in the direction a positive charge would accelerate by convention).

A negative charge in the same field would accelerate the opposite way. Imagine an object that would do that in a gravitational field! Where does such an object have 0 PE? Really far away. So the amount of potential energy is the SAME, but it has a different SIGN meaning whether or not you need to work WITH the field (-) or AGAINST (+) the field to move that charge.

Finally, look at the equation for potential energy between two charges.

`U=(kQq)/r`

if the charges attract, the potential energy is negative. It will require work to *separate* the charges.

if the charges repel, the potential energy is positive suggesting that it will take work to position the charges *closer* together.

Notice, though, that the magnitude is the same.