what does the negative sign shows in electric potential gradient?

1 Answer

valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

We will take a day by day example for clarity: the electric force and the electric potential.

Suppose you have a charge +Q that creates an electric field `E` whose value is

`E(R) = k_e*Q/R^2`

The force of a test charge +q is repelling and has a value of

`F = q*E = k_e*(qQ)/R^2`

Now we want to compute the work that one person need to do to move the test charge from a distance `R` from to infinity in the field `E` . Because the force is repelling, the work will be negative (the field is doing work on the test charge, not the person moving it).

`W = -int_R^(+oo)F(R)dR =k_e*(qQ)/R (R->+oo) =-k_e*(qQ)/R`

Now we DEFINE the electric potential as being the work done by a person to move the charge from `R` to `+oo` over the value of the test charge

`U = W/q = -k_e*Q/R`

Thus the potential `U` associated with the field at a distance `R` from the charge is

`U = -int_0^RE(R)*dR (= -k_e*Q/R)`

which means

`E = -gradU`   or equivalent

`F =-qgradU`

Therefore the force is always opposing to changes in potential values (or in other words, positive test charges always moves towards smaller potentials- it is like a mass always fall from a higher hill position towards the base of the hill).

Now you have the answer to the question: the negative sign shows that the force of a field is directed in such a way as to make the variation of potential smaller.