Electric potential
| English | Chinese | Pinyin |
|---|---|---|
| electric potential | 电势 | diàn shì |
| scalar | 标量 | biāoliàng |
| electric potential energy | 电势能 | diàn shì néng |
| infinity | 无穷远 | wú qióng yuǎn |
| potential gradient | 电势梯度 | diàn shì tī dù |
| bound | 束缚 | shù fù |
The graph that answers four questions at once
- A favourite exam question draws the electric potential along the line between two charged spheres and asks four things: the signs of the charges, where the field is zero, the ratio of the charges, and the force on a charge placed at a point.
- All four come from the same curve, because potential is a scalar that simply adds, and the field is the gradient of the graph.
- Learn to read that one picture and a whole class of questions becomes routine.
- This lesson is electric potential 电势, its link to the field, and electric potential energy 电势能.
Defining potential
- The electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity 无穷远 to that point:
- Both marks: work done per unit positive charge, and from infinity to the point. The potential is zero at infinity, which is what makes the definition well posed.
- For a positive source charge $V > 0$ everywhere outside it, because work must be done on a positive test charge to push it in against the repulsion. For a negative source charge $V < 0$.
Electric potential at a point is the work per unit charge to bring a positive charge from:
Like gravitational potential, it is measured from infinity, where the potential is taken as zero.
The two-mark definition of electric potential needs which elements? Select all that apply.
Per unit positive charge, and from infinity. The zero at infinity is what makes the definition well posed; stationarity is not part of it.
Potential of a point charge
- Note the $\dfrac{1}{r}$, where the field goes as $\dfrac{1}{r^2}$. That difference is the most-tested fact in the topic.
- $V$ is a scalar 标量. For several charges, add the potentials with their signs: no directions, no resolving, just a sum.
A curve above the axis, or below it, depending only on the sign of Q
Electric potential
V = kQ / r
Potential ∝ 1/r around a charge — steep near it, flattening out.
The electric potential near a point charge varies as 1/r.
Yes — $V = \dfrac{Q}{4\pi\varepsilon_0 r}$ goes as $1/r$, while the field goes as $1/r^{2}$.
To find the potential at a point due to several charges, add the individual potentials with their signs.
Potential is a scalar, so there is nothing to resolve. Fields, by contrast, must be added as vectors with directions.
Field is the negative potential gradient
- The field at a point equals minus the potential gradient 电势梯度 there:
- The minus sign says the field points towards lower potential: a positive charge released in the field runs "downhill" in $V$.
- In a uniform field the potential falls steadily, so the gradient is constant and this reduces to $E = V/d$, which is the parallel-plate result from earlier.
- Practically: to get the field from a $V$ against $x$ graph, draw a tangent and take minus its gradient.
The electric field equals minus the potential ____ (E = −dV/dx).
The field points "downhill" in potential, toward lower V.
How is the electric field found from a graph of potential against distance?
E = -dV/dx. The minus sign says the field points towards lower potential, and in a uniform field the constant gradient gives E = V/d.
Worked example: reading the two-sphere graph
- The potential along the line between spheres X and Y, $1.2\ \text{m}$ apart, is positive everywhere and has a minimum at $0.50\ \text{m}$ from X. State three conclusions.
- Both charges are positive, because the potential is positive everywhere along the line.
- The field is zero at $x = 0.50\ \text{m}$, because the gradient is zero there and $E = -dV/dx$.
- Y carries the larger charge, because the zero-field point lies closer to X. Quantitatively $Q_{\text{Y}}/Q_{\text{X}} = (0.70/0.50)^2 = 2.0$.
- If instead the curve crossed zero, the charges would be opposite, and there the ratio comes from $Q_{\text{X}}/x = Q_{\text{Y}}/(d-x)$, without the square.
The potential between two spheres is positive everywhere with a minimum nearer sphere X. Which conclusions follow? Select all that apply.
A minimum in V is where the gradient, and so the field, is zero; it lies nearer the smaller charge. The potential there is a minimum, not zero, so qV is not zero either.
Electric potential energy
- A charge $q$ placed at a point of potential $V$ has electric potential energy $E_{\text{P}} = qV$. For two point charges:
- Like charges give positive potential energy: released, they fly apart. Opposite charges give negative energy, a bound 束缚 system, and work must be done to separate them to infinity.
- This is the same pattern as gravitational potential energy, which is always negative because gravity always attracts.
Above the axis they would fly apart; below it they are bound
The electric potential energy of a charge $q$ at a point of potential $V$ is:
$E_{\text{P}} = qV$; for two point charges this becomes $\dfrac{Qq}{4\pi\varepsilon_0 r}$.
Two opposite charges have a negative electric potential energy (a bound system).
Negative PE means energy must be supplied to pull them apart — they are bound, like an electron and a nucleus.
Two opposite charges have a negative electric potential energy, so the system is described as ____.
Work must be done to separate them to infinity, where the energy is zero. Like charges have positive energy and would fly apart if released.
Electric and gravitational, side by side
| gravitational | electric | |
|---|---|---|
| force | $F = \dfrac{Gm_1m_2}{r^2}$ | $F = \dfrac{Q_1Q_2}{4\pi\varepsilon_0 r^2}$ |
| field | $g = \dfrac{GM}{r^2}$ | $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$ |
| potential | $\phi = -\dfrac{GM}{r}$ | $V = \dfrac{Q}{4\pi\varepsilon_0 r}$ |
- Identical shapes throughout. The one difference: mass has a single sign so gravity always attracts and $\phi$ is always negative, while charge has two signs so the electric quantities can be either.
Match each quantity to how it changes with distance from a point charge.
The field is inverse-square; the potential is inverse (one power of r gentler).
Worked example: describe the motion
- A positive charge is placed at a point P on the two-sphere line and released. Describe and explain its motion.
- It moves towards lower potential, that is, down the slope of the $V$ graph, because the field points that way and the force on a positive charge is along the field.
- The force, and so the acceleration, is not constant: it equals $-q\,dV/dx$, and the gradient changes along the line.
- It speeds up wherever it moves to lower potential, gaining kinetic energy equal to $q\Delta V$ lost in potential energy.
Marks that slip away
- Potential is $\dfrac{1}{r}$; field is $\dfrac{1}{r^2}$. Check which the question wants before substituting.
- The definition needs per unit positive charge and from infinity. Zero at infinity is what the definition rests on.
- Potentials add as scalars with signs; fields add as vectors. Never resolve potentials.
- Zero field is where the $V$ graph has zero gradient, not where $V$ itself is zero. Between opposite charges the curve crosses zero while its gradient is not zero.
You've got it
- electric potential is the work done per unit positive charge bringing a small test charge from infinity; it is zero at infinity, positive near a positive charge
- $V = \dfrac{Q}{4\pi\varepsilon_0 r}$ varies as $\dfrac{1}{r}$ and is a scalar, so several charges add with signs
- $E = -\dfrac{dV}{dx}$: the field is minus the potential gradient, so zero field means zero gradient, and in a uniform field this gives $E = V/d$
- electric potential energy $E_{\text{P}} = qV = \dfrac{Qq}{4\pi\varepsilon_0 r}$: positive for like charges, negative for opposite ones, which are bound