Conservation of Electric Energy
Introduced| English | Meaning |
|---|---|
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | The energy an object has because of its motion. |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | The energy transferred per unit charge between two points. |
| electron-volt/ɪˈlektrɒn vəʊlt/ | electron-volt |
Let a charge go and stored energy becomes speed
- Release a charge in a field and it accelerates on its own.
- Its electric potential energy falls while its kinetic energy 动能 rises.
- The total energy stays the same — nothing is lost.
- This one idea solves problems that would be hard with forces alone.
Energy is conserved
- With only the electric force acting: $\Delta KE + \Delta U = 0$.
- Energy just changes form, from potential to kinetic (or back).
- So a gain in speed is paid for by a drop in potential energy.
- No need to track the changing force along the path — just the endpoints.
A charge is released in a field (only the electric force acts). As it speeds up, its potential energy:
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$\Delta KE + \Delta U = 0$: as KE rises, $U$ falls.
Charge falling through a voltage
- Moving a charge $q$ through a potential difference 电势差 $\Delta V$ changes its energy by $q\,\Delta V$.
- If that is the only force: $q\,\Delta V = \Delta KE = \tfrac12 m v^2$ (from rest).
- A $+$ charge speeds up moving to lower potential; a $-$ charge speeds up moving to higher.
- The sign of the charge decides which way is "downhill".
A $2\ \text{C}$ charge falls through $\Delta V = 5\ \text{V}$. How much kinetic energy (in J) does it gain?
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$KE = q\,\Delta V = 2 \times 5 = 10\ \text{J}$.
An electron speeds up when it moves toward:
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A negative charge gains KE moving to higher $V$ (its $q\Delta V < 0$ means $\Delta U < 0$).
A handy unit: the electron-volt
- One electron-volt 电子伏特 (eV) is the energy an electron gains crossing $1\ \text{V}$.
- $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ — tiny, but perfect for particles.
- Accelerators quote energies in keV, MeV, and GeV.
- It saves writing awkward powers of ten for single charges.
Speeding up or slowing down?
As a charge moves through a field, energy converts between kinetic and potential. Sort each case.
The energy an electron gains crossing $1\ \text{V}$ is one ____.
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That energy is $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.
Path doesn't matter
- The energy change depends only on the start and end potentials.
- Any route between them gives the same $q\,\Delta V$.
- The electric force is conservative — like gravity.
- That is exactly why a potential (a single number per point) can exist.
The energy change $q\Delta V$ depends only on the start and end potentials, not the path.
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The electric force is conservative — path-independent.
Select all true statements about electric energy conservation.
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Total energy is conserved, the force is conservative, eV is energy. Path does not matter.
An electron (charge $e$) is accelerated from rest through $100\ \text{V}$. Find its kinetic energy.
- $KE = q\,\Delta V = e \times 100\ \text{V} = 100\ \text{eV}$.
- In joules: $100 \times 1.6\times10^{-19} = 1.6\times10^{-17}\ \text{J}$.
Watch the sign of the charge. A positive charge speeds up going to lower potential, but a negative charge (like an electron) speeds up going to higher potential. The energy $q\,\Delta V$ carries that sign automatically.
With only the electric force, energy is conserved: $\Delta KE + \Delta U = 0$. A charge through $\Delta V$ gains $q\,\Delta V$ of energy, so $q\,\Delta V = \tfrac12 m v^2$ from rest. The force is conservative (path-independent); a handy unit is the electron-volt.
Write from an energy gain of 20 microjoules
- A charge of 4 microcoulomb starts at rest and moves through a potential drop of 5 volts.
- Find the energy it gains. Say whether a longer path changes that energy.
A charge of 4 microcoulomb moves through a potential drop of 5 volts. How much energy does it gain, and does a longer path change that energy?
4 times 10 to the minus 6 coulomb times 5 volts is 20 times 10 to the minus 6 joules.
How much energy does this charge gain?
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4 microcoulomb times 5 volts is 20 microjoules, for any path between those potentials.