# Trivial but heavy doubt in Potential

## Main Question or Discussion Point

Trivial but heavy doubt in Potential!!

I am having a trivial but heavy confusion with negative confusion of negative sign in the calculation of the potential.When we bring the charge from infinity to radial distance r,the electrical field and displacement are in opposite direction but we just multiply them without taking direction into consideration.Why and How??

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mukundpa
Homework Helper
The external force applied to bring the charge is equal to electric force in magnitude but opposite in direction.

quasar987
Homework Helper
Gold Member
Right; since displacement and electric force are of opposite sign, we must do positive work to bring them closer (i.e. the force we use to bring them closer will be in the same direction of the displacement). And since $W = -\Delta V$, the potential energy of the system is negative. Does this answer your confusion?

lightgrav
Homework Helper
Here's a different perspective:
the PE is the Work that the E-field *would* do (potentially)
if the charge were taken *from* that place *to* infinity.

Isn't that why it's called "Potential" Energy?

Potential is just the PE of a "unit" (positive) test charge.
Then, E is parallel dr (for positive source Q), so PE and V are positive.

If conditions change with time, the "Work done coming in from infinity"
no longer describes the Energy available in the system
(that could be released on the way out to infinity, or to somewhere else).

(By the way, Quasar left out a "q" in W = - Delta V)

quasar987
Homework Helper
Gold Member
I have the bad habit (curse you Symon!) to call the potential energy V.

It's too late to correct my post.

ehild
Homework Helper
lightgrav said:
Here's a different perspective:
the PE is the Work that the E-field *would* do (potentially)
if the charge were taken *from* that place *to* infinity.

Potential is just the PE of a "unit" (positive) test charge.
Then, E is parallel dr (for positive source Q), so PE and V are positive.
Lightgrav is right, the statement often cited that "the potential is the work of an external force needed to bring a unit positive charge to the place from infinity" is a bit misleading. What is the potental around a negative point charge? No force is needed to bring a positive charge from infinity to a distance r. It goes by itself. Is the potential zero then? The definition using external force would be correct only if it contained the condition that the KE of the unit positive charge should not change.

The work done by the electric field when a charge moves form point A to B is

$$W= \int_A^B{\vec{E}\cdot d \vec {r} =U_A-U_B=-\Delta U$$

as the potential difference is defined as the final potential minus initial potential.

$$\Delta U=U_B-U_A$$

If B is at infinity, and the potential is zero there, the work is $U_A$.

The potential around a positive charge is positive: the field repels the positive test charge, both force and displacement point outward, the work of the field is positive when the test charge moves to infinity. Around a negative charge, there is a backward force exerted by the field when the test charge moves away to infinity, so the work of the field is negative and so is the potential.

ehild