Electric field of a point charge
Introduced| English | Meaning |
|---|---|
| point charge/pɔɪnt tʃɑːdʒ/ | An idealised electric charge treated as concentrated at a single point. |
| inverse-square/ɪnˈvɜːs skweə/ | Describing a quantity that varies in proportion to one divided by the square of another quantity. |
| vector sum/ˈvektə sʌm/ | The resultant obtained by adding vectors while accounting for their magnitudes and directions. |
Take away the second charge
- Coulomb's law needs two charges. But a charge sitting alone still changes the space around it, whether or not anything is there to feel it.
- Divide the Coulomb force by the test charge and the test charge cancels out. What is left describes the source charge alone: its field.
- That is the whole move from force to field, and it is the reason a field is worth defining at all.
- This lesson is the field of a point charge 点电荷, its inverse-square fall, and how fields from several charges combine.
The field of a point charge
- Take Coulomb's law and divide by the test charge, $E = F/q$:
- It points outwards from a positive $Q$ and inwards towards a negative one, and it falls as $\dfrac{1}{r^2}$.
- Only the source charge appears. That is the point: the field exists whether or not another charge is there to feel it.
Radial lines, and a curve that drops away fast
Field of a point charge
E ∝ Q/r²
The electric field of a point charge spreads out radially and obeys the inverse-square law.
The field strength at distance $r$ from a point charge $Q$ is:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
From $E = F/q$ with Coulomb's force, the test charge cancels, leaving $E = \dfrac{Q}{4\pi\varepsilon_0 r^{2}}$.
The field is $100\ \dfrac{\text{N}}{\text{C}}$ at distance $r$ from a charge. What is it at $2r$?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Inverse-square: $\dfrac{100}{2^{2}} = 25\ \dfrac{\text{N}}{\text{C}}$.
The field points ____ from a positive point charge.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A positive charge pushes a positive test charge away, so its field points outward.
Exactly like gravity
- Compare $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$ with $g = \dfrac{GM}{r^2}$. The same inverse-square 平方反比 shape, with $Q$ in place of $M$ and $\dfrac{1}{4\pi\varepsilon_0}$ in place of $G$.
- The one real difference: charge has two signs, so an electric field can push or pull, while a gravitational field only ever pulls.
- Because the shapes match, every result transfers. Halve the distance and both fields quadruple.
The electric field of a point charge falls off with distance just like gravity (as 1/r²).
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Both are inverse-square: $E = \dfrac{Q}{4\pi\varepsilon_0 r^{2}}$ and $g = \dfrac{GM}{r^{2}}$.
How do the electric field and the electric potential of a point charge depend on distance?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Confusing these two is the commonest error in the topic. The field is the negative gradient of the potential, which is exactly why one power lower appears in V.
Worked example: the charged Earth
- Treat the Earth as a conducting sphere of radius $6.37 \times 10^6\ \text{m}$ carrying $-4.80 \times 10^5\ \text{C}$. Find the field at the surface and compare it with $g$.
- Outside a sphere the charge acts as a point charge at the centre, so $r$ is the radius:
- It is directed towards the centre, because the charge is negative. Gravity gives $g = GM/R^2 = 9.83\ \text{N/kg}$, so $E/g \approx 11$.
- Say the direction as well as the size. For a negative source charge the field points inwards, and half the mark is there.
A sphere of radius 6.37e6 m carries -4.80e5 C. What is the field strength at its surface, in V/m? (1/4 pi eps0 = 8.99e9)
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Outside a sphere the charge acts as a point charge at the centre, so r is the radius: E = 8.99e9 x 4.80e5 / (6.37e6)^2 = 106 V/m, directed towards the centre.
Several charges: add as vectors
- The total field at a point is the vector sum 矢量和 of the fields of each charge separately.
- Add them as arrows, with size and direction, not as numbers. Two fields of $100\ \text{V/m}$ in opposite directions give zero, not two hundred.
- This is where the electric field differs from electric potential, which is a scalar and adds with sign alone. Keep the two operations apart.
For several charges, the total field at a point is:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Fields are vectors, so add them with both size and direction.
Worked example: where is the field zero?
- Point charges of $+4.0\ \text{nC}$ and $+1.0\ \text{nC}$ are $30\ \text{cm}$ apart. Where on the line joining them is the resultant field zero?
- Between the charges the two fields point in opposite directions, so they can cancel there. Let $x$ be the distance from the $4.0\ \text{nC}$ charge:
- The zero lies twice as far from the larger charge, which is the sanity check: the bigger field needs more distance to weaken.
- For opposite charges the two fields between them point the same way and never cancel, so the zero lies outside, beyond the smaller charge.
Charges of +4.0 nC and +1.0 nC are 0.30 m apart. How far from the 4.0 nC charge, in metres, is the resultant field zero?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Set the magnitudes equal: 4/x^2 = 1/(0.30-x)^2, so (0.30-x)/x = 1/2 and x = 0.20 m, twice as far from the larger charge.
For two opposite point charges, there is a point between them where the resultant field is zero.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Between opposite charges the two fields point the same way and add, so they never cancel. The zero lies outside, beyond the smaller charge.
Reading a field-distance graph
- For a charged conducting sphere, $E$ is zero inside, jumps to its maximum at the surface, and then falls as $1/r^2$ outside.
- The radius is the distance at which the field jumps from zero, and the charge follows from any point on the curve using $Q = 4\pi\varepsilon_0 r^2 E$.
- Given the surface value $E_0$, the field is $E_0/4$ at $2R$ and $E_0/9$ at $3R$. Sketch through those points rather than guessing the curve.
On a graph of field strength against distance from the centre of a charged conducting sphere, which are true? Select all that apply.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
The radius is where the field jumps from zero, and the charge follows from any point on the outside curve using Q = 4 pi eps0 r^2 E.
Marks that slip away
- Field is $\dfrac{1}{r^2}$; potential is $\dfrac{1}{r}$. Confusing them is the single commonest error in this topic.
- Give the direction: outwards from positive, inwards towards negative.
- Fields add as vectors; potentials add as scalars. Two equal opposite fields cancel.
- For like charges the zero field lies between them; for opposite charges it lies outside, beyond the smaller one.
You've got it
- dividing Coulomb's law by the test charge leaves the source charge alone: $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$, outwards from positive and inwards towards negative
- it is inverse-square, the same shape as $g = GM/r^2$, but charge has two signs so the field can repel
- several charges add as a vector sum, unlike potentials which are scalars
- the zero-field point lies between like charges, nearer the smaller one, and outside opposite charges, beyond the smaller one