Electric Potential Energy
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| work/wɜːk/ | công |
| electric potential energy/ɪˈlektrɪk pəˈtenʃl ˈenədʒi/ | thế năng điện |
Squeeze two like charges together and you bottle up energy
- Push two like charges closer and the field fights you every step.
- The work 功 you do doesn't vanish — it is stored.
- Release the charges and that stored energy pushes them apart, gaining speed.
- Stored energy that depends only on position is potential energy.
Energy to assemble charges
- Electric potential energy 电势能 $U$ is the work to bring the charges from far away to here.
- For two point charges: $U = \dfrac{k\,q_1 q_2}{r}$.
- Note it is $1/r$, not $1/r^2$ — energy, not force.
- $U$ is a scalar (just a number with a sign), so it is easier than force.
The potential energy of two point charges is proportional to:
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$U = kq_1q_2/r$ — energy goes as $1/r$ (force goes as $1/r^2$).
Find $U$ (in J) for a $+1\ \text{nC}$ and $+1\ \text{nC}$ charge $1\ \text{m}$ apart. Use $k = 9\times10^9$.
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$U = \dfrac{(9\times10^9)(10^{-9})(10^{-9})}{1} = 9\times10^{-9}\ \text{J}$.
The sign tells the story
- Like charges: $U > 0$ — you had to push them together, energy is stored.
- Opposite charges: $U < 0$ — they pulled together, energy is "owed" (bound).
- A more negative $U$ means a more tightly bound pair.
- We choose $U = 0$ when the charges are infinitely far apart.
Two opposite charges have a potential energy that is:
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Opposite charges attract, so $U = kq_1q_2/r < 0$ — a bound pair.
We usually set the potential energy to zero when the charges are infinitely ____.
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The reference $U = 0$ is taken at infinite separation.
Work and force from energy
- The work done by the field equals the drop in energy: $W_{\text{field}} = -\Delta U$.
- So moving to lower $U$ releases energy as motion.
- In calculus form the force is the slope of $U$: $F = -\dfrac{dU}{dr}$.
- Energy and force are two views of the same interaction.
Electric potential energy
The potential energy of two charges falls off as one over the distance between them.
The work done by the electric field equals the decrease in potential energy, $W = -\Delta U$.
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A drop in $U$ releases energy as work: $W = -\Delta U$.
Many charges add up
- For several charges, add the energy of every pair.
- Because $U$ is a scalar, you just add numbers — no vectors.
- $U_{\text{total}} = \sum_{\text{pairs}} \dfrac{k\,q_i q_j}{r_{ij}}$.
- Keep the signs: a mix of $+$ and $-$ can give a net negative energy.
Select all true statements about electric potential energy.
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$U$ is a signed scalar, summed over pairs; like charges give $U>0$. It is never a vector.
How much energy to bring a $+2\ \text{nC}$ charge to $3\ \text{cm}$ from a $+3\ \text{nC}$ charge?
- $U = \dfrac{k\,q_1 q_2}{r} = \dfrac{(9\times10^9)(2\times10^{-9})(3\times10^{-9})}{0.03}$.
- $U \approx 1.8\times10^{-6}\ \text{J}$ — positive, since both are positive.
$U$ is a scalar with a sign, not a force and not a vector. A negative $U$ doesn't mean a negative force — it means the pair is bound (energy would be needed to pull them apart).
Electric potential energy is the work to assemble charges: $U = k q_1 q_2 / r$ (a signed scalar, $1/r$). Like charges store $U>0$; opposite charges give $U<0$ (bound). The field's work is $W = -\Delta U$, and $F = -dU/dr$.
Write from a potential energy of 6 joules
- Charges of +20 microcoulomb and +10 microcoulomb are 0.30 m apart. Use k = 9 times 10 to the 9.
- Find the potential energy and its sign.
Find the electric potential energy of a +20 microcoulomb charge and a +10 microcoulomb charge separated by 0.30 m. Use k = 9 times 10 to the 9.
9 times 10 to the 9 times 20 times 10 times 10 to the minus 12, divided by 0.30, is 6 joules.
What is the potential energy of this pair?
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1.8 divided by 0.30 is 6 joules, and like charges store positive energy.