Electric Field Strength
The Charge & Coulomb's Law topic described the force between two charges as if each one reached across empty space and pushed or pulled the other. This topic describes it in two steps. A charge changes the space around it, whether or not anything is there to feel it, and the changed space is called an electric field (电场). A second charge placed in the field feels a force from the field at the point where it sits. Splitting the force this way separates two questions: how strong the field is at a point, and what force it gives to whatever charge is put there. This page answers both. It defines the electric field strength, E = F/q, turns it round into F = qE, and uses Coulomb's law to find the field of a single point charge, E = kQ/r², which falls with the square of the distance.
Three of the four papers so far have asked a one-step question about field strength, each at Q28 or Q29: the force on a charge in a given field (December 2025, Q29), the ratio of the fields at distances r and 3r from a charge (March 2026, Q29), and the field after the distance doubles (April 2026, Q28). The papers tend to keep a question type in the same place, so expect a field-strength item near there again. All three are worked through below. Each takes a line or two once the ideas are clear, and their wrong options come from a short list of slips: a power of ten lost in the arithmetic, a distance that was not squared, and a ratio written upside down.
The field and the test charge
Fix a small charged ball on an insulating stand, and call its charge Q. To find out what it does to the space around it, you need something that responds: a second, much smaller charge q, called a test charge (试探电荷). Put the test charge at a point P and it feels a force. Move it to another point and the force has a different size, and perhaps a different direction. A force is waiting at every point around the ball, so the space itself has been changed. That changed space is the electric field of Q, and Q is called the source charge (场源电荷) of the field.
The force on its own is a poor measure of the field, because it depends on the test charge as well. Coulomb's law shows how. With Q and P fixed, the force F = kQq/r² is proportional to q: a test charge twice as large feels twice the force, and one three times as large feels three times the force. So the ratio F/q comes out the same whichever test charge you use. The same is true near any arrangement of charges, since each charge's push on the test charge is proportional to q. The ratio depends on the source charges and on where P is, and on nothing else, so it describes the field at P. It is called the electric field strength at P.
You have met this idea before, for gravity. In the Weight, Springs & Friction topic, g = 10 N/kg is the force the Earth exerts on each kilogram near its surface. A 2 kg bag feels 20 N and a 5 kg bag 50 N, but both work out at the same 10 N for each kilogram. g describes the Earth's pull at that place, whatever is put there to feel it. The electric field strength is the same idea for charge: the force for each coulomb.
E = F/q is a definition by ratio (比值定义), like v = Δx/Δt or R = U/I, which the Unit Conversions in Problems page listed beside it. A formula of this kind tells you how to measure a quantity, not what the quantity depends on. Reading it as "E is proportional to F" or "E is inversely proportional to q" is wrong: change q and F changes in step, so E stays the same. Take the test charge away and the force is gone, but the field is not. E at the point is the same with or without a charge there to feel it.
A test charge needs to be small in two ways. It must be small in size, so that it measures the field at one point and not across a region. And its charge must be small, so that its own pull does not shift the source charges and change the very field it is measuring.
Which way the field points, and F = qE
A force has a direction, so the field has one too: E is a vector. December 2025, Q1 listed electric field strength among its vectors, as the Vectors & Scalars page showed. Its direction is fixed by agreement: the field at a point points the way the force on a positive charge placed there points.
Coulomb's law then gives the direction around a point charge. A positive test charge near a positive source charge is repelled, so the field of +Q points directly away from it. A positive test charge near a negative source charge is attracted, so the field of −Q points directly towards it. Either way the field lies along the line from the source charge to the point.
Now turn the definition round. Multiplying E = F/q by q gives F = qE, the force on any charge q at a point where the field strength is E. A positive charge is pushed along the field. A negative charge feels a force of the same size, |q|E, in the opposite direction. At a single point the field has a single direction, yet charges of opposite sign placed there are pushed opposite ways.
A point charge of 3.0 × 10⁻⁹ C is placed in an electric field of 6.0 × 10⁴ N/C. It experiences an electric force of ( ).
A. 1.8 × 10⁵ N B. 1.8 × 10⁻⁴ N C. 2.0 × 10⁻⁴ N D. 1.8 × 10⁻⁵ N
"An electric field of 6.0 × 10⁴ N/C" is the papers' way of giving the field strength at the place where the charge sits. Both values are already in SI units, so F = qE applies straight away.
Front numbers: 3.0 × 6.0 = 18. Powers of ten: 10⁻⁹ × 10⁴ = 10⁻⁵. So F = 18 × 10⁻⁵ N = 1.8 × 10⁻⁴ N. The answer is B.
The question gives no sign and no direction, so it asks only for the size. Had the charge been −3.0 × 10⁻⁹ C, the size would be the same and the force would point against the field.
D, 1.8 × 10⁻⁵ N, is the tidy-up slip: the front number went from 18 to 1.8 while the exponent stayed at −5, which makes the force ten times too small. A, 1.8 × 10⁵ N, has the right front number with a positive power of ten: about the weight of 18 tonnes, from a few billionths of a coulomb. C, 2.0 × 10⁻⁴ N, has the right power of ten but divides the front numbers, 6.0 ÷ 3.0, instead of multiplying them. The Unit Conversions in Problems page used this same question to practise powers of ten.
An electron, of charge −1.6 × 10⁻¹⁹ C, is at a point where the electric field is 5.0 × 10³ N/C, pointing east. Find the force on it.
Step 1, the size: F = |q|E = 1.6 × 10⁻¹⁹ × 5.0 × 10³. The front numbers give 1.6 × 5.0 = 8.0 and the powers 10⁻¹⁹ × 10³ = 10⁻¹⁶, so F = 8.0 × 10⁻¹⁶ N.
Step 2, the direction: the electron is negative, so the force points against the field, to the west.
A proton at the same point, with charge +1.6 × 10⁻¹⁹ C, would feel a force of the same size, 8.0 × 10⁻¹⁶ N, to the east. The field at the point is one thing; the force depends on what is put there.
The field of a point charge
For a single point charge, Coulomb's law gives the field directly. Put a test charge q at a distance r from a point charge Q. The force on it is F = kQq/r². Divide by the test charge, as the definition says:
E = F/q = (kQq/r²) ÷ q = kQ/r².
The test charge cancels, as it has to. The field at a point is set by the source charge Q and by the distance r from it, which is what the graph above showed: P′, twice as far from the source charge as P, had a quarter of the field strength. The direction comes from the last section: away from Q if Q is positive, towards Q if Q is negative.
Keep the two formulas apart. E = F/q is the definition, and it holds in every field, whatever makes it. E = kQ/r² is the field of one point charge only, and in it Q is always the source charge, never the test charge. When several charges make the field, each contributes its own kQ/r² at the point, and the next page adds them; December 2025, Q38 and April 2026, Q38 do exactly that at the midpoint between two charges.
With numbers, most of the work is in the units and the powers of ten. Convert the charge to coulombs and the distance to metres before you square it, as the Unit Conversions in Problems page practised.
A point charge Q = +2.0 × 10⁻⁸ C is fixed in a vacuum, and point P is 20 cm from it. Find the field at P, and the force on a charge of −2.0 × 10⁻⁹ C placed at P. (k = 9.0 × 10⁹ N·m²/C²)
Step 1, SI units: r = 20 cm = 0.20 m, so r² = 0.040 m² = 4.0 × 10⁻² m².
Step 2, the field: kQ = 9.0 × 10⁹ × 2.0 × 10⁻⁸ = 18 × 10¹ = 1.8 × 10². Then E = 1.8 × 10² ÷ (4.0 × 10⁻²) = 0.45 × 10⁴ = 4.5 × 10³ N/C. Q is positive, so the field at P points away from Q.
Step 3, the force: F = |q|E = 2.0 × 10⁻⁹ × 4.5 × 10³ = 9.0 × 10⁻⁶ N. The charge at P is negative, so the force points against the field: towards Q.
Check with Coulomb's law in one go: kQq/r² = 9.0 × 10⁹ × 2.0 × 10⁻⁸ × 2.0 × 10⁻⁹ ÷ 0.040 = 3.6 × 10⁻⁷ ÷ 0.040 = 9.0 × 10⁻⁶ N, and unlike charges attract. The field route gives the same answer in two steps, and its first step is reusable: any charge placed at P feels |q| × 4.5 × 10³ N/C.
How the field changes with distance
Most questions about E = kQ/r² give no numbers at all. They call the field at one point E, change the distance or the charge, and ask for the new field. The ratio method of the Coulomb's Law page works unchanged. Write the new field over the old one, and k cancels:
E′/E = (kQ′/r′²) ÷ (kQ/r²) = (Q′/Q) × (r/r′)².
The source charge brings its factor just as it is. The distance brings its factor squared and turned upside down, because r sits squared in the bottom of the formula. So twice the distance gives a quarter of the field, three times the distance a ninth, and half the distance four times the field.
| Distance from Q | Field strength | Distance factor |
|---|---|---|
| r | E | ×1 |
| 2r | E/4 | ×1/2² = ×1/4 |
| 3r | E/9 | ×1/3² = ×1/9 |
| 4r | E/16 | ×1/4² = ×1/16 |
| r/2 | 4E | ×2² = ×4 |
| r/3 | 9E | ×3² = ×9 |
To compare two points in the field of one charge, write out both fields and divide:
EA : EB = kQ/rA² : kQ/rB² = rB² : rA².
The distances swap sides. That swap is easy to lose, so check the answer against the picture: the nearer point always has the stronger field.
In vacuum, points A and B are at distances r and 3r from charge Q respectively. The ratio of their electric field strengths is ( ).
A. 1 : 9 B. 3 : 1 C. 9 : 1 D. 1 : 3
Step 1, both fields: EA = kQ/r² and EB = kQ/(3r)² = kQ/(9r²).
Step 2, the ratio: EA : EB = kQ/r² : kQ/(9r²). The kQ cancels, and multiplying both sides by 9r² leaves 9 : 1. The answer is C.
Check: A is the nearer point, so its field must be the stronger, and the ratio EA : EB must be greater than 1. That rules out A and D at once.
A, 1 : 9, has the right numbers the wrong way round: it is EB : EA. B, 3 : 1, forgets to square the distance, as if the field fell as 1/r. D, 1 : 3, makes both slips.
A point charge Q produces an electric field strength E at point A. If the distance from the point charge to point A doubles, the electric field strength becomes ( )
A. E/2 B. 2E C. E/4 D. 4E
Step 1, the charge: Q does not change, so its factor is ×1.
Step 2, the distance: it doubles, so its factor is ×1/2² = ×1/4.
Step 3: E′ = E × 1 × 1/4 = E/4. The answer is C.
A, E/2, forgets the square. D, 4E, squares the factor but turns it the wrong way up: 4E is the field at half the distance. B, 2E, makes the field grow in step with the distance, which gets both the square and the direction wrong; moving away from a charge always weakens its field.
When the source charge changes as well, multiply its factor in. The two changes can pull in opposite directions, and the factors settle which one wins.
A point charge Q produces a field of strength E at a certain point. The charge is halved, and the field is now measured at a point only a third as far from it. Find the new field strength.
Step 1, the charge: halved, so ×1/2.
Step 2, the distance: r′ = r/3, so (r/r′)² = 3² and the factor is ×9.
Step 3: E′/E = 1/2 × 9 = 9/2, so E′ = 4.5E.
Halving the charge halves the field, but moving three times closer makes it nine times stronger, and the distance wins.
One more distinction keeps these questions straight. The test charge changes the force, never the field. Moving to a new point changes E, through r. Changing the test charge changes only F, through F = qE. A question that changes both needs both steps, in that order: first the field at the new point, then the force on the new charge.
Your turn! · Practice questions
Things to remember
| Concept | What to know |
|---|---|
| Electric field | A charge changes the space around it. A charge placed there feels a force from the field at its own position. The source charge Q makes the field; a small test charge q measures it. |
| Field strength | E = F/q, in N/C: the force for each coulomb. It belongs to the point. Change the test charge and F changes while E does not; take it away and the field is still there. |
| Definition by ratio | E = F/q says how to measure E, not what it depends on. "E ∝ F" and "E ∝ 1/q" are false. Compare g = 10 N/kg, the force for each kilogram. |
| Direction | E is a vector and points along the force on a positive charge. Around a point charge it points away from +Q and towards −Q. |
| F = qE | Size |q|E. Along E for a positive charge, against E for a negative one. December 2025 Q29: 3.0 × 10⁻⁹ × 6.0 × 10⁴ = 1.8 × 10⁻⁴ N. |
| Point charge | E = kQ/r², from Coulomb's law with the test charge cancelled; k = 9.0 × 10⁹ N·m²/C². Q is the source charge in C, r the distance from it in m. E = F/q holds in every field; E = kQ/r² only for a point charge. |
| Inverse square | r, 2r, 3r, 4r give E, E/4, E/9, E/16; r/2 gives 4E. The nearer point has the stronger field. March 2026 Q29: 9 : 1. April 2026 Q28: E/4. |
| Ratios | E′/E = (Q′/Q) × (r/r′)², and k cancels. For two points of one charge, EA : EB = rB² : rA². |
| Powers of ten | Multiply: add the exponents. Divide: subtract them. Then tidy the front number to between 1 and 10. Convert cm to m before squaring. |
| Common slips | 1/r for 1/r² (3 : 1, E/2); the ratio upside down (1 : 9); E divided by q instead of multiplied; the tidy-up slip (1.8 × 10⁻⁵ N); a distance left in centimetres; a test charge thought to change the field. |