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We usually say that the work done by a force is $\mathrm{d}W = F\cdot\mathrm{d}l$, and the electric potential is $\mathrm{d}U = -E \cdot \mathrm{d}l$). Why do we put a negative sign over there? Is it due to some convention for the E-Field?

Qmechanic
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Nasir Mahmood
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  • You do external force on the object in field which is opposite to E. Since it is opposite hence -E – Shashaank Mar 01 '17 at 10:40
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    "Why is there a minus in the definition* of the electric potential?*" There is no "why" for definitions. – lucas Mar 01 '17 at 14:11
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    @lucas I disagree. Definitions can't be "wrong," but they can more more or less useful or well-motivated. Understanding the motivation for a definition usually goes a long way in helping to understand the underlying physics. – tparker Jun 16 '17 at 21:58
  • See https://physics.stackexchange.com/questions/64674/potential-energy-sign-conventions – ProfRob Apr 30 '20 at 07:48

3 Answers3

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A good way to think of this issue is to image the potential energy of a rollercoaster (i.e., the height of the rollercoaster track above the ground). The force on the cart at any given point on the track is equal the negative slope of the track at that point. When the cart falling down a steep downward incline, the slope of the track is negative, so the cart is experiences an acceleration. When cart is going up a hill, the slope is positive, so the cart experiences a deceleration (negative acceleration).

More succinctly, forces act to equalize energy. If there exists a gradient in the energy of a given system, there will be an associated force that acts to remove this gradient. Since this force is acting to $remove$ the gradient in the energy, it comes with a $minus sign$.

Thus, $F = -\nabla U$.

Matt P.
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It is because while deriving electric potential we consider the work done by external force and not the electric field.

The fact to be noted here is that while external force is doing work on a charged particle the kinetic energy of particle is $0$, which means that the energy that the charged particle possesses is only potential energy.

So, if external force is either greater or smaller than the force due to electric field, there will be net force that will accelerate the charged particle. We do not want that to happen since we are only interested in the finding the electric potential (work done per unit charge against electric field) and if we provide more external force then we will to do more work.

We just want to consider the work done against the electric field and not the additional work done in increasing the kinetic energy.

Therefore, the external force is equal to force due to electric field in magnitude. As, we want no net force on the charged particle $(\approx 0)$ the external force has to be opposite to the force due to electric field.

$$\vec F_{ext}=-q\vec{E}$$

So you can just replace $\vec F_{ext}$ by $-q\vec{E}$ when calculating the work done by external force against electric field.

Hence the electric potential, the work done by $-q \vec{E}$, is

$$\Delta V := \int (-q \vec{E}) \cdot d\vec{l} = -q \int \vec{E} \cdot d\vec{l}.$$

We can see from the above integral that the negative sign naturally follows.

abouttostart
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Mitchell
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It's because people like having the total energy kinetic + potential to be constant over time. If potential energy were defined with the opposite sign, then the energy difference kinetic - potential would be conserved instead, which seems less natural.

tparker
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