Redistribution of Charge Between Conductors
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| grounding/ˈɡraʊndɪŋ/ | nối đất |
| surface charge density/ˈsɜːfɪs tʃɑːdʒ ˈdensɪti/ | mật độ điện tích mặt |
Connect a big sphere to a small one — who gets more charge?
- Join two charged conductors with a wire and charge suddenly flows.
- It keeps flowing until something between them is equal.
- That something is potential, not charge — a subtle but key point.
- Get this right and you understand grounding, sharing, and lightning rods.
They end at the same potential
- Charge flows from high potential to low, just like water finding its level.
- It stops when both conductors sit at the same $V$.
- At that moment no potential difference remains to push charge.
- Equal potential — not equal charge — is the end state.
When two conductors are connected, charge flows until they have equal:
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Charge flows until the potentials are equal — no push remains.
Bigger conductors hold more charge
- For a sphere, $V = \dfrac{kQ}{R}$, so equal $V$ means $Q \propto R$.
- The bigger sphere ends up with the larger share of charge.
- Double the radius and it holds double the charge at the same potential.
- Size sets how much charge a conductor can hold per volt.
At equal potential, the smaller sphere holds more charge.
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False. $V = kQ/R$ equal means $Q \propto R$, so the larger sphere holds more charge (although the smaller one has the stronger surface field).
A radius-$3R$ sphere shares charge with a radius-$R$ sphere. The big one holds how many times the charge of the small one?
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$Q \propto R$, so the $3R$ sphere holds $3\times$ the charge.
But the small one has denser charge
- Surface charge density 面电荷密度 is $\sigma \propto \dfrac{1}{R}$ — higher on the smaller sphere.
- So the small, sharp conductor has the stronger surface field.
- That is the lightning-rod effect again: sharp $=$ dense $=$ strong field.
- Big total charge and high density live on different conductors.
Sharing charge between conductors
When two conductors are connected, charge flows until they reach the same potential.
Which conductor has the higher surface charge density?
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$\sigma \propto 1/R$, so the smaller conductor is denser.
Select all true statements about connected conductors.
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Equal potential, $Q \propto R$, $\sigma \propto 1/R$. The split is by size, not 50/50.
Grounding is sharing with the Earth
- Grounding 接地 connects a conductor to the enormous Earth.
- The Earth is so big it holds any charge at essentially $V = 0$.
- Charge flows until the conductor also reaches $V = 0$.
- That is how induction charging "drains" the unwanted sign.
Grounding connects a conductor to the Earth, pulling it to a potential of ____.
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The Earth holds any charge at $V = 0$.
A sphere of radius $2R$ is joined to one of radius $R$, sharing charge.
- Equal potential: $Q \propto R$, so the big sphere holds twice the charge.
- But $\sigma \propto 1/R$, so the small sphere has the denser surface charge.
Connected conductors share to equal potential, not equal charge. Don't split the charge 50/50 — split it in proportion to size ($Q \propto R$ for spheres), and remember density runs the other way ($\sigma \propto 1/R$).
Connected conductors exchange charge until they reach the same potential. For spheres, $Q \propto R$ (the bigger holds more), while surface density $\sigma \propto 1/R$ (the smaller is denser). Grounding shares with the Earth, pulling a conductor to $V = 0$.
Write from a 6 and 2 microcoulomb split
- A sphere of radius 3R is joined by a wire to a sphere of radius R. Together they hold 8 microcoulomb.
- Find the charge on each sphere after they reach the same potential.
Spheres of radius 3R and R share 8 microcoulomb until their potentials are equal. How much charge does each hold, and which has the larger surface density?
Equal potential means Q is proportional to R. Of 8 microcoulomb, three quarters is 6 and one quarter is 2.
How do these spheres share 8 microcoulomb?
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The ratio of radii is 3 to 1, so the charges are 6 and 2 microcoulomb. Density goes the other way.