Electric Fields
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| field/fiːld/ | trường |
| electric field/ɪˈlektrɪk fiːld/ | điện trường |
| test charge/test tʃɑːdʒ/ | điện tích thử |
| uniform/ˈjuːnɪfɔːm/ | đều |
How does one charge feel another across empty space?
- A charge pushes another far away, with nothing visibly between them.
- We explain it with a field 场: charge fills the space around it with influence.
- Any other charge placed there feels a force from the local field.
- The field is the middle-man that carries the electric force.
Field = force per unit charge
- The electric field 电场 at a point is the force a $+1\ \text{C}$ test charge 检验电荷 would feel there.
- $\vec E = \dfrac{\vec F}{q}$, measured in newtons per coulomb (N/C).
- $E$ is a vector: it has a direction at every point.
- Once you know $E$, the force on any charge is just $\vec F = q\vec E$.
The electric field at a point is the:
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$\vec E = \vec F / q$ — the force a unit positive charge would feel.
Once you know the field $E$, the force on a charge $q$ is $F =$ ____.
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$\vec F = q\vec E$ — multiply the field by the charge.
The field of a point charge
- A point charge $Q$ makes a field $E = \dfrac{kQ}{r^2}$ at distance $r$.
- It points away from a positive $Q$ and toward a negative $Q$.
- It is strongest near the charge and fades as $1/r^2$.
- Field lines show the direction; where they crowd, the field is stronger.
Map a charge's field
Change the sign and size of the charge and watch how its field lines rearrange.
Find the field (in N/C) $3\ \text{m}$ from a $+6\ \text{nC}$ charge. Use $k = 9\times10^9$.
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$E = \dfrac{(9\times10^9)(6\times10^{-9})}{3^2} = \dfrac{54}{9} = 6\ \text{N/C}$.
Reading field lines
- Lines start on positive charge and end on negative charge.
- They never cross — the field has one direction at each point.
- Closer lines mean a stronger field.
- The line's tangent gives the direction of $\vec E$ (and the force on a $+$ charge).
Two electric field lines can cross at a point.
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Lines never cross — the field has only one direction at each point.
Select all true statements about field lines.
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Lines run + to −, crowd where strong, and never cross.
A uniform field
- Between two large parallel plates the field is uniform 均匀 — same everywhere.
- The lines are straight, evenly spaced, and point from $+$ to $-$.
- A uniform field gives a constant force, so a charge moves like a projectile.
- We will use this constant-field picture for capacitors.
Between two large parallel plates the field is:
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Parallel plates make a uniform field: straight, evenly spaced lines.
Find the field $2\ \text{cm}$ from a $+5\ \text{nC}$ charge. Use $k = 9\times10^9$.
- $E = \dfrac{kQ}{r^2} = \dfrac{(9\times10^9)(5\times10^{-9})}{(0.02)^2}$.
- $E \approx 1.1\times10^{5}\ \text{N/C}$, pointing away from the charge.
The field $\vec E$ exists whether or not a test charge is there — it is a property of space made by the source charge. Don't say "no charge, no field": the source still fills space with $E$.
The electric field is the force per unit charge, $\vec E = \vec F / q$ (N/C), a vector pointing away from $+$ and toward $-$. A point charge makes $E = kQ/r^2$; field lines show its direction, and closer lines mean a stronger field.
Write from a field of 1800 newtons per coulomb
- A +8 nanocoulomb point charge is the source. The point is 0.20 m away. Use k = 9 times 10 to the 9.
- Find the field and its direction.
Find the electric field 0.20 m from a +8 nanocoulomb charge. Use k = 9 times 10 to the 9. Include the direction.
9 times 10 to the 9 times 8 times 10 to the minus 9, divided by 0.04, is 1800. Field direction is the force on a positive test charge.
What is the field 0.20 m from this charge?
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kQ over r squared is 72 divided by 0.04, which is 1800 newtons per coulomb, pointing away.