Electric Potential
Introduced| English | Meaning |
|---|---|
| volt/vəʊlt/ | The SI unit of potential difference, equal to one joule of energy transferred per coulomb of charge. |
| electric potential/ɪˈlektrɪk pəˈtenʃl/ | Electric potential energy per unit positive charge at a point, relative to a chosen reference. |
| equipotential/ˌiːkwɪpəˈtenʃl/ | A surface or set of points that all have the same electric or gravitational potential. |
| gradient/ˈɡreɪdɪənt/ | The spatial rate and direction of change of a quantity; for a scalar field its gradient points towards the steepest increase. |
The "height" of the electric landscape
- Imagine the field as a hilly landscape and charge as a ball.
- Potential tells you the "height" at each point — energy per unit charge.
- A $+$ charge rolls downhill (to low potential); a $-$ charge rolls uphill.
- Because it is energy per charge, it is a scalar — no arrows to juggle.
Volts: energy per charge
- The electric potential 电势 $V$ is the potential energy per unit charge: $V = \dfrac{U}{q}$.
- Its unit is the volt 伏特 (1 V $= 1$ J/C).
- For a point charge: $V = \dfrac{kQ}{r}$ — positive near $+$, negative near $-$.
- The energy of a charge placed there is simply $U = qV$.
Electric potential $V$ is:
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$V = U/q$ — potential energy per unit charge, in volts.
Scalars add the easy way
- Potential is a scalar, so many charges just add their $V$ values.
- $V = \sum \dfrac{kQ_i}{r_i}$ — no components, no angles.
- This is why potential is often easier to compute than the field.
- Find $V$ everywhere first, then get the field from it.
Potentials from several charges must be added as vectors, using their directions.
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False. Potential is a scalar: $V = \sum kQ_i/r_i$, so you add the values with their signs. It is the electric field that must be added as vectors.
Two charges give $+50\ \text{V}$ and $-20\ \text{V}$ at a point. Find the total $V$ (in V).
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Scalars add: $50 + (-20) = 30\ \text{V}$.
Equipotentials map the field
- An equipotential 等势面 is a line (or surface) of constant $V$.
- No work is done moving a charge along an equipotential.
- Equipotentials are always perpendicular to field lines.
- For a point charge they are circles; for a uniform field, parallel lines.
Field and equipotentials
See how the field lines from a charge cross its circular equipotentials at right angles.
Equipotential surfaces are always ____ to the electric field lines.
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Field lines cross equipotentials at right angles.
Field is the slope of potential
- The field points downhill, from high $V$ to low $V$.
- Its size is the steepness: $E = -\dfrac{dV}{dx}$ (a gradient 梯度).
- Closely packed equipotentials mean a steep slope — a strong field.
- So $E$ (a vector) and $V$ (a scalar) carry the same information.
Where equipotentials are packed closely together, the field is:
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A steep potential slope ($E = -dV/dx$) means a strong field.
Select all true statements about $V$ and $E$.
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$V$ scalar, $E$ vector, $E = -dV/dx$. But $V=0$ does NOT force $E=0$.
Two charges give $V = +30\ \text{V}$ and $V = -12\ \text{V}$ at a point. What is the total potential?
- Potential is a scalar, so just add: $V = 30 + (-12)$.
- $V = +18\ \text{V}$ — no vectors needed.
$V$ is a scalar; $E$ is a vector. They are not the same: at the midpoint between two equal opposite charges, $V = 0$ but $E$ is not zero. Never assume "$V=0$ means $E=0$".
Electric potential is energy per charge, $V = U/q$ (volts), a scalar you can add: $V = \sum kQ_i/r_i$. Equipotentials are constant-$V$ surfaces, perpendicular to the field, and the field is the slope $E = -dV/dx$.
Write from a potential of 32 volts
- One charge contributes +40 volts at a point. Another contributes -8 volts there.
- Find the total potential. Say why the fields are not added in the same way.
At one point two charges contribute +40 volts and -8 volts. What is the total potential, and why is the field addition different?
40 plus -8 is 32. Scalars add directly. Fields need components.
What is the total potential?
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40 plus -8 is 32 volts.