Electric field strength E equals F over q and E equals V over d — Physics

Before this

This leads to

Electric potential and potential energy · coming soon
Motion of charges in electric fields · coming soon

Two ways to measure the same invisible thing

An electric field cannot be seen, weighed or touched. So how do you say how strong it is?

The first lecture in this topic gave one answer. Put a small positive test charge in the field, measure the force on it, and divide by the charge. That is a measurement made with forces. But there is a second answer, and it comes from a completely different place. Connect two metal plates to a battery, measure the potential difference between them with a voltmeter, measure the gap with a ruler, and divide. No force is measured at all.

Both methods give the same number, in what look like two different units. This lecture shows why. On the way, it finishes the job the last lecture started: the field of a single charge, worked out in detail, and the uniform field between plates. By the end, you will also understand the experiment that first weighed the charge of one electron.

The definition: force per unit charge

The field strength E at a point is the force per unit positive charge on a small test charge placed there:

E = \frac{F}{q}

F is the force in newtons and q is the test charge in coulombs, so E is measured in newtons per coulomb, N/C. This is a definition. It does not depend on what made the field, so it works everywhere: near a single charge, between plates, or near a charged cloud.

Turn it round and you get the equation you will use most often: $F = qE$ . A charge of 2 μC in a field of 5000 N/C feels a force of $2 \times 1 0^{- 6} \times 5000 = 0.01$ N. The direction comes from the sign. A positive charge is pushed along the field. A negative charge, such as an electron, is pushed the opposite way.

The field of a single charge

The last lecture gave the force between a source charge Q and a test charge q at a distance r: $F = k Qq / r^{2}$ . Now divide by q, as the definition says:

E = \frac{F}{q} = \frac{k Qq}{r ^{2} q} = \frac{k Q}{r ^{2}}

Look at what just happened. The test charge q cancels. The field depends only on the source charge Q and the distance r. It does not depend on the charge you use to measure it. This is why we can say that the field belongs to the space around Q, whether or not anything is there to feel it.

Because r is squared, the field falls fast. Double the distance and E becomes a quarter. Triple it and E becomes a ninth. This is the inverse-square rule again: the field lines spread out over a sphere whose area grows as r².

Field of a point charge

At a distance r from a point charge Q, the field strength is $E = \frac{k Q}{r ^{2}}$ , with $k \approx 8.99 \times 1 0^{9} N m^{2} / C^{2}$ .

The field points away from a positive charge and towards a negative charge. The same rule holds outside a charged sphere, with r measured from its centre.

The uniform field between plates

Now a very different arrangement. Two flat metal plates face each other, a distance d apart, and a battery holds a potential difference V between them. One plate becomes positive and the other negative.

Between parallel plates the field lines are straight, parallel and evenly spaced, so the field is uniform. Only near the edges do the lines bulge outwards (grey). A small positive charge feels a force F = qE towards the negative plate.

The field lines run straight across from the positive plate to the negative plate. They are parallel, and they are evenly spaced. Evenly spaced lines mean the same field strength everywhere between the plates, away from the edges. This is called a uniform field. Near the edges the lines bulge outwards and the field weakens. In most problems the plates are wide compared with the gap, so the edges can be ignored.

Before you read on, test your picture of a uniform field.

Predict, then reveal — where is the force largest?

A small positive charge sits between two parallel plates, far from the edges. You move it from just below the positive plate, to the middle, to just above the negative plate. What happens to the force on it? Choose an answer before you read on.

The same small positive charge is shown at three places between the plates, away from the edges. The force on it is the same at all three, because the field is uniform: the lines are parallel and evenly spaced.

The same charge near the positive plate, in the middle and near the negative plate. The force arrows are equal.

This is the most useful property of the uniform field. A charge feels the same force wherever it is between the plates. That is exactly like gravity near the ground, where a stone feels the same weight at every height you are likely to meet.

Why E = V/d: follow the energy

Here is the second way to measure the field. It needs one fact you already know from circuits. The potential difference between two points is the energy transferred per coulomb of charge that moves between them: $V = W / q$ . One volt is one joule per coulomb.

This is the hard part of the lecture. Spend time here. Follow a small positive charge q as the field carries it from the positive plate to the negative plate.

Step 1. The force on the charge is $F = qE$ . In a uniform field, this force is the same all the way across.
Step 2. The charge moves a distance d along the force. Work done = force × distance, so the field does work $W = qE \times d$ .
Step 3. By the meaning of potential difference, the energy transferred to a charge q crossing a p.d. V is $W = q V$ .
Step 4. These are the same energy, so $qE d = q V$ . Divide both sides by qd.

E = \frac{V}{d}

That is it. That is the whole idea. The force view says how hard the field pushes. The energy view says how much energy it hands over. Push × distance = energy, so the field is the energy per coulomb, spread over the gap.

Notice where Step 1 used the word "uniform". The derivation needs the same force all the way across. So $E = V / d$ is true only for a uniform field. Near a single charge, the force changes with distance, and this short argument fails.

Uniform field between parallel plates

For plates a distance d apart with a potential difference V between them, the field strength between the plates, away from the edges, is $E = \frac{V}{d}$ .

The field points from the positive plate to the negative plate.

The equation also says what changes the field. Raise the p.d. and the field rises in proportion. Move the plates apart, with the same p.d., and the field falls: the same energy per coulomb is now spread over a longer distance.

Stop and think

Two plates are 2.0 cm apart, connected to a 100 V supply. They are moved to 4.0 cm apart, still connected. What happens to the field, and to the force on a charge between them?

Show the answer

The p.d. stays at 100 V, because the supply holds it there. The gap doubles, so $E = V / d$ halves: from 5000 V/m to 2500 V/m. The force $F = qE$ halves too. A common wrong answer is "no change, because the voltage is the same". The field depends on the p.d. per metre, not on the p.d. alone.

One quantity, two units

From $E = F / q$ , the unit of field strength is N/C. From $E = V / d$ , it is V/m. These must be the same unit, and they are. Follow the definitions:

\frac{V}{m} = \frac{J / C}{m} = \frac{N m / C}{m} = \frac{N}{C}

A volt is a joule per coulomb, and a joule is a newton metre. The metres cancel. So 5000 V/m and 5000 N/C are exactly the same field.

Top: near a point charge, the field falls with distance as 1/r², so at 2r it is a quarter of its value at r. Bottom: between parallel plates, the field is the same all the way across the gap.

The two graphs show why you must choose the right equation. $E = k Q / r^{2}$ describes a field that changes from point to point. $E = V / d$ describes a field that does not. Neither can replace the other. Only $E = F / q$ works for both, because it is the definition.

Weighing the charge of one electron

A uniform field gives the same upward force at every height. That made possible one of the most famous experiments in physics. In 1909, at the University of Chicago, Robert Millikan began measuring the charge on tiny charged droplets. Water drops evaporated too quickly, so his student Harvey Fletcher used drops of oil from a simple perfume sprayer.

The drops fell between two horizontal plates. With the field switched off, a drop's falling speed gave its size, and so its mass. With the field switched on, the p.d. was adjusted until the electric force upwards balanced the weight downwards, and the drop hung still. Then:

qE = m g so q = \frac{m g}{E} = \frac{m g d}{V}

Millikan found that the charge on every drop was a whole-number multiple of one small charge. He reported its value in 1913 as $1.592 \times 1 0^{- 19}$ C. The modern value is $1.602 \times 1 0^{- 19}$ C. The small difference came mainly from the value he used for the viscosity of air. He received the Nobel Prize in 1923, partly for this work. Try a version of the experiment below.

Real-world model — hold an oil drop still

Choose a drop. Its mass has already been measured. The plates are 5.0 mm apart, with the positive plate at the top, so a negatively charged drop is pulled upwards. Adjust the p.d. until the drop hangs still. Then read off its charge. (The drops are invented, with g = 9.8 m/s².)

Drop:

p.d. between the plates: 300 V

This table shows the p.d. that holds each invented drop still between plates 5.0 mm apart, and the charge it gives by q = mgd/V, with g = 9.8 m/s² and e = 1.60 × 10⁻¹⁹ C.

Drop	m / 10⁻¹⁵ kg	p.d. / V	q / 10⁻¹⁹ C	q ÷ e
A	3.2	490	3.20	2
B	2.4	245	4.80	3
C	5.6	≈ 429	≈ 6.40	4

Each charge is a whole number of electron charges. A real drop's mass is found from how fast it falls with the field off; this model gives it to you.

Each drop hangs still at a different p.d., because each has a different mass and charge. But every charge comes out as 2, 3 or 4 times the same small number. That is what Millikan saw, drop after drop: charge comes in whole units, and the unit is the charge of one electron.

Worked examples

Worked Example 1 — Plates and a charge

Two parallel plates are 4.0 cm apart, with a p.d. of 200 V between them. (a) Find the field strength between the plates. (b) Find the force on a charge of $+ 3.0 μ C$ placed between them.

Think first. Parallel plates mean a uniform field, so use $E = V / d$ . Convert the gap to metres first.

(a) $d = 4.0 cm = 0.040 m$ . So $E = 200 \div 0.040 = 5000 V/m$ , from the positive plate towards the negative plate.

(b) $F = qE = 3.0 \times 1 0^{- 6} \times 5000 = 1.5 \times 1 0^{- 2} N$ , towards the negative plate, because the charge is positive.

Check. Units: V/m × C = N/C × C = N. ✓ And the size is sensible: 0.015 N is roughly the weight of a 1.5 g coin, small but real.

Worked Example 2 — The field inside an atom

A proton is $2.0 \times 1 0^{- 10}$ m from an electron. Find the field strength at the proton due to the electron, and the force on the proton. Use $k = 9.0 \times 1 0^{9} N m^{2} / C^{2}$ and $e = 1.6 \times 1 0^{- 19}$ C.

Think first. The electron is a point charge, so use $E = k Q / r^{2}$ with sizes only. Find the direction afterwards.

E = \frac{9.0 \times 1 0 ^{9} \times 1.6 \times 1 0 ^{- 19}}{( 2.0 \times 1 0 ^{- 10} ) ^{2}}

The top is $1.44 \times 1 0^{- 9}$ . The bottom is $4.0 \times 1 0^{- 20}$ . So $E = 3.6 \times 1 0^{10} N/C$ . The electron is negative, so the field at the proton points towards the electron.

Force on the proton: $F = qE = 1.6 \times 1 0^{- 19} \times 3.6 \times 1 0^{10} = 5.76 \times 1 0^{- 9} N \approx 5.8 \times 1 0^{- 9} N$ , towards the electron.

Check. Coulomb's law directly: $F = k e^{2} / r^{2} = 9.0 \times 1 0^{9} \times (1.6 \times 1 0^{- 19})^{2} \div (2.0 \times 1 0^{- 10})^{2} = 5.76 \times 1 0^{- 9}$ N. ✓ Two routes, one answer. A field of 36 thousand million volts per metre is far beyond anything made in a laboratory between plates. Fields this strong exist only on the scale of atoms.

Worked Example 3 — An electron between plates

An electron is between plates 2.0 cm apart with a p.d. of 500 V. Find the force on it, and its acceleration. Compare the force with its weight. (Electron mass $9.11 \times 1 0^{- 31}$ kg.)

Think first. Field from $E = V / d$ , force from $F = qE$ , acceleration from $a = F / m$ .

$E = 500 \div 0.020 = 25000 V/m$ .

$F = 1.6 \times 1 0^{- 19} \times 25000 = 4.0 \times 1 0^{- 15} N$ , towards the positive plate, because the electron is negative.

$a = F / m = 4.0 \times 1 0^{- 15} \div 9.11 \times 1 0^{- 31} \approx 4.4 \times 1 0^{15} m/s^{2}$ .

Weight: $m g = 9.11 \times 1 0^{- 31} \times 9.8 \approx 8.9 \times 1 0^{- 30}$ N. The electric force is about $4.5 \times 1 0^{14}$ times larger.

Check. The ratio $a / g = 4.4 \times 1 0^{15} \div 9.8 \approx 4.5 \times 1 0^{14}$ is the same as the force ratio, as it must be. ✓ This is why electron problems ignore gravity: it is far too small to matter.

Worked Example 4 — A drop held still

An oil drop of mass $4.9 \times 1 0^{- 15}$ kg hangs still between horizontal plates 5.0 mm apart when the p.d. is 500 V. Find the charge on the drop, and how many electron charges it carries. (Numbers chosen for practice; g = 9.8 m/s².)

Think first. "Hangs still" means the forces balance: $qE = m g$ , with $E = V / d$ .

q = \frac{m g d}{V}

Substitute: the top is $m g d = 4.9 \times 1 0^{- 15} \times 9.8 \times 5.0 \times 1 0^{- 3} \approx 2.40 \times 1 0^{- 16}$ . Divide by 500, so $q \approx 4.80 \times 1 0^{- 19}$ C. Dividing by $1.60 \times 1 0^{- 19}$ gives about 3.0. The drop carries three electron charges.

Check. $E = 500 \div 0.0050 = 1.0 \times 1 0^{5}$ V/m, and $qE = 4.80 \times 1 0^{- 19} \times 1.0 \times 1 0^{5} = 4.80 \times 1 0^{- 14}$ N, while $m g = 4.9 \times 1 0^{- 15} \times 9.8 \approx 4.80 \times 1 0^{- 14}$ N. ✓ The forces balance.

Three traps

Trap 1 — The wrong equation for the field

$E = V / d$ is only for a uniform field, such as between parallel plates. $E = k Q / r^{2}$ is only for a point charge or outside a charged sphere. Read the situation first, then choose. When in doubt, $E = F / q$ is always true.

Trap 2 — Centimetres and millimetres

A gap of 4.0 cm is 0.040 m, and 5.0 mm is 0.0050 m. Put centimetres into $E = V / d$ and the answer is 100 times too small. Write the conversion on its own line before you substitute.

Trap 3 — The direction for a negative charge

The field direction is the direction of the force on a positive charge. An electron between plates is pushed towards the positive plate, against the field lines. Work out the size with $F = qE$ , then decide the direction from the sign.

Try it yourself

Work each one on paper first. Use $k = 9.0 \times 1 0^{9} N m^{2} / C^{2}$ , $e = 1.6 \times 1 0^{- 19}$ C and g = 9.8 m/s². Open a hint only when you are stuck.

Two plates are 1.5 cm apart with a p.d. of 6.0 kV. Find the field strength, and the force on a charge of 2.0 nC between them.

Hint 1
Convert first: 6.0 kV = 6000 V, 1.5 cm = 0.015 m, 2.0 nC = 2.0 × 10⁻⁹ C.

Hint 2
Use $E = V / d$ , then $F = qE$ .

Full answer
$E = 6000 \div 0.015 = 4.0 \times 1 0^{5} V/m$ . $F = 2.0 \times 1 0^{- 9} \times 4.0 \times 1 0^{5} = 8.0 \times 1 0^{- 4} N$ . Check: V/m × C = N, and doubling the charge would double the force, as $F = qE$ says. ✓
A small sphere carries a charge of $+ 5.0 μ C$ . Find the field strength 30 cm from its centre, and 60 cm from its centre.

Hint 1
Use $E = k Q / r^{2}$ with r in metres.

Hint 2
For the second distance, you can use the inverse-square rule instead of calculating again.

Full answer
$E = 9.0 \times 1 0^{9} \times 5.0 \times 1 0^{- 6} \div 0.3 0^{2} = 4.5 \times 1 0^{4} \div 0.090 = 5.0 \times 1 0^{5} N/C$ , pointing away from the sphere. At 60 cm the distance has doubled, so the field is a quarter: $1.25 \times 1 0^{5} N/C$ . Check: $4.5 \times 1 0^{4} \div 0.36 = 1.25 \times 1 0^{5}$ . ✓
An oil drop of mass $1.6 \times 1 0^{- 15}$ kg carries two electron charges. The plates are 8.0 mm apart. What p.d. holds the drop still?

Hint 1
Still means $qE = m g$ , and $E = V / d$ .

Hint 2
Rearrange to $V = m g d / q$ , with $q = 2 \times 1.6 \times 1 0^{- 19}$ C.

Full answer
$q = 3.2 \times 1 0^{- 19}$ C. $V = m g d / q = 1.6 \times 1 0^{- 15} \times 9.8 \times 0.0080 \div 3.2 \times 1 0^{- 19} \approx 392 V$ . Check: $E = 392 \div 0.0080 = 4.9 \times 1 0^{4}$ V/m, and $qE = 3.2 \times 1 0^{- 19} \times 4.9 \times 1 0^{4} \approx 1.57 \times 1 0^{- 14}$ N, equal to $m g = 1.6 \times 1 0^{- 15} \times 9.8 \approx 1.57 \times 1 0^{- 14}$ N. ✓
An electron starts at rest at the negative plate of a pair of plates 1.0 cm apart, with a p.d. of 100 V. (a) Find its acceleration. (b) Use $v^{2} = u^{2} + 2 a s$ to find its speed when it reaches the positive plate. (c) Show that the answer does not depend on the gap, by writing the speed in terms of e, V and m. (Electron mass $9.11 \times 1 0^{- 31}$ kg.)

Hint 1
$E = V / d$ , $F = e E$ , $a = F / m$ . The distance s is the gap d.

Hint 2
For (c), substitute $a = e V / (m d)$ and $s = d$ into $v^{2} = 2 a s$ . What happens to d?

Full answer
(a) $E = 100 \div 0.010 = 1.0 \times 1 0^{4}$ V/m, $F = 1.6 \times 1 0^{- 19} \times 1.0 \times 1 0^{4} = 1.6 \times 1 0^{- 15}$ N, so $a = 1.6 \times 1 0^{- 15} \div 9.11 \times 1 0^{- 31} \approx 1.76 \times 1 0^{15} m/s^{2}$ . (b) $v^{2} = 2 \times 1.76 \times 1 0^{15} \times 0.010 \approx 3.51 \times 1 0^{13}$ , so $v \approx 5.9 \times 1 0^{6}$ m/s. (c) $v^{2} = 2 \times \frac{e V}{m d} \times d = \frac{2 e V}{m}$ . The d cancels, so $v = 2 e V / m$ . Check: $2 \times 1.6 \times 1 0^{- 19} \times 100 \div 9.11 \times 1 0^{- 31} \approx 5.9 \times 1 0^{6}$ m/s. ✓ The speed depends only on the p.d., because the energy gained is $e V$ , whatever the gap. The next lecture builds on exactly this idea.

The bigger idea

Put the two equations side by side with gravity. Near the ground, the gravitational field g is uniform, a mass feels a force mg, and lifting it through a height h stores energy mgh. Between plates, the electric field E is uniform, a charge feels a force qE, and moving it across the gap transfers energy qV = qEd. The same pattern appears twice: field × distance = energy per unit charge, or per unit mass.

So $E = V / d$ is a statement about how fast the energy per coulomb changes as you move across the field. Where the p.d. changes quickly with distance, the field is strong. The field also points from high potential to low potential, the way a ball rolls downhill. The next lecture makes this exact. It defines the potential at every point in a field, even near a single charge where $E = V / d$ no longer holds.

Key concepts in this lesson

Electric field strength is defined as force per unit positive charge, E = F/q, measured in N/C, and this definition works for every field.
For a point charge, E = kQ/r², which depends only on the source charge and the distance, and falls to a quarter when the distance doubles.
Between parallel plates, away from the edges, the field is uniform, and E = V/d, which follows from the work done by the field equalling qV.
The units N/C and V/m are the same, because a volt is a joule per coulomb and a joule is a newton metre.
A uniform field gives the same force at every point, which let Millikan balance charged oil drops and show that charge comes in whole units of about 1.6 × 10⁻¹⁹ C.