Uniform electric field
Why do the dust particles of an air purifier end up on the surface of the plates of the device, although the air flows past the plates?
Prior knowledge
- Coulomb's law: the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distancechapter Electric charge and Coulomb's law
- Voltage tells the energy per unit charge: $1\,\text{V} = 1\,\text{J/C}$chapter Electric circuit, current and voltage
- Newton's second law $\overline{F} = m\overline{a}$chapter Newton's second law
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
An electric air purifier sucks room air into itself. At the start of the device the dust particles get an electric charge, and then the air flows between two parallel metal plates with a voltage of several kilovolts between them. The air comes out clean, and the dust stays as a grey layer on the surface of the plates, although the plates are not in the way of the airflow.
Why do the charged dust particles drift to the surface of the plates, and does the force on a particle depend on where between the plates it is?
Optional Review: Coulomb's law
Predict
Make your predictions before investigating 2 questions
Prediction 1: the force at different places
Prediction 2: the distance of the plates
Investigate
Adjust the voltage and the distance of the plates, and the charge and position of the ball. Read the reading of the force sensor. At which places is the force the same, and what determines its magnitude?
Simulation
How to use
The sliders set the voltage $U$ between the plates, the distance $d$ of the plates, the charge $q$ of the ball and the position of the ball: the horizontal position $x$ from the middle of the plates and the height $h$ from the lower plate. The plates are $20\,\text{cm}$ long, so their ends are at $x = -10\,\text{cm}$ and $x = 10\,\text{cm}$. The upper plate is connected to the positive terminal of the voltage source. The ball is attached to a force sensor, which shows the electric force $F$ on the ball and its direction. The red arrow is the same force, and its length grows with the force. Field shows the green field lines, the field vectors, or neither. The number of field lines grows with the electric field strength, and the length of a field vector tells the electric field strength at that point. The signs on the plates show their charge.
Assumptions of the model
The plates are thin conductors whose charge settles so that each plate is at the same potential throughout. Around the plates there is vacuum or air, and the charge of the ball itself does not change the charges of the plates. The model is two-dimensional: the plates continue very far in the direction perpendicular to the plane of the picture. The force sensor measures only the electric force, so the weight of the ball is not shown. The numbers are values calculated by the model, not measurements.
How the calculation works
The field is not set by the equation $E = U : d$. Each plate is divided into $40$ strips, and the charges of the strips are solved from a system of equations so that the potential of the upper plate is $+U : 2$ and that of the lower plate $-U : 2$. The field at any point is the vector sum of the fields of all the strips. From the electric field strength $\overline{E}$ calculated at the ball, the force of the sensor is obtained as
$$\overline{F} = q\overline{E}$$
The field lines are drawn by moving in small steps always in the direction of the field, starting from the surface of the positive plate.
Experiment 1: the force at different heights
Test prediction 1. The voltage $2.0\,\text{kV}$, the plate distance $4.0\,\text{cm}$, the charge of the ball $+2.0\,\text{nC}$ and the horizontal position $x = 0$ stay constant; change only the height $h$ of the ball. Record the force at no fewer than three heights, for example $0.5\,\text{cm}$, $2.0\,\text{cm}$ and $3.5\,\text{cm}$. How does the force depend on the height?
Initial state: voltage $U$ = $2\,\text{kV}$, plate distance $d$ = $4\,\text{cm}$, charge of the ball $q$ = $2\,\text{nC}$, horizontal position $x$ = $0\,\text{cm}$, height above the lower plate $h$ = $0{.}5\,\text{cm}$, field: field lines shown. Kept constant: voltage $U$, plate distance $d$, charge of the ball $q$, horizontal position $x$. Varied: height above the lower plate $h$.
| height above the lower plate $h$ | $F$ $(\mu\text{N})$ |
|---|
Experiment 2: force and charge
The voltage $2.0\,\text{kV}$, the plate distance $4.0\,\text{cm}$ and the position of the ball stay constant; change only the charge $q$ of the ball. Record the force with no fewer than three charges, including a negative one. How do the force and its direction depend on the charge?
Initial state: voltage $U$ = $2\,\text{kV}$, plate distance $d$ = $4\,\text{cm}$, charge of the ball $q$ = $1\,\text{nC}$, horizontal position $x$ = $0\,\text{cm}$, height above the lower plate $h$ = $2\,\text{cm}$, field: field lines shown. Kept constant: voltage $U$, plate distance $d$, horizontal position $x$, height above the lower plate $h$. Varied: charge of the ball $q$.
| charge of the ball $q$ | $F$ $(\mu\text{N})$ |
|---|
From your table, calculate the electric field strength $E = F : q$ at the ball. Answer in the unit $\text{N/C}$.
Hint 1: a guiding question
Divide the force of each row by the magnitude of the charge in the same row. What do you notice?
Hint 2: what is the connection
First convert the units: $1\,\mu\text{N} = 10^{-6}\,\text{N}$ and $1\,\text{nC} = 10^{-9}\,\text{C}$.
Hint 3: an intermediate step
For example, the row $q = 3\,\text{nC}$ gives $E = 150 \cdot 10^{-6}\,\text{N} : (3 \cdot 10^{-9}\,\text{C})$.
Show the solution
Given
readings of the model at the same place, for example with the charge $q = 3.0\,\text{nC}$ the force $F = 150\,\mu\text{N}$ and with the charge $q = 1.0\,\text{nC}$ the force $F = 50\,\mu\text{N}$
Find
the electric field strength $E$ in the unit $\text{N/C}$
Principle
The ball is at the same place in every row of the table, and the field of the plates is the same. The force is directly proportional to the charge, so the quotient of the force and the charge is the same in every row. This quotient is exactly the electric field strength at the place.
Formula
$E = \dfrac{F}{q}$
Conversions
Substitution
Answer
$5.0 \cdot 10^{4}\,\text{N/C}$
Two significant figures like the charge.
Reasonableness
The row $q = 1.0\,\text{nC}$ gives the same:
With a negative charge the force is equally large but points up, against the field. The value is the same as $U : d = 2000\,\text{V} : 0.040\,\text{m}$.
Experiment 3: the distance of the plates
Test prediction 2. The voltage $2.0\,\text{kV}$, the charge of the ball $+2.0\,\text{nC}$, the horizontal position $x = 0$ and the height $h = 1.0\,\text{cm}$ stay constant; change only the distance $d$ of the plates. Record the force at no fewer than three distances. How does the force depend on the distance of the plates?
Initial state: voltage $U$ = $2\,\text{kV}$, plate distance $d$ = $2\,\text{cm}$, charge of the ball $q$ = $2\,\text{nC}$, horizontal position $x$ = $0\,\text{cm}$, height above the lower plate $h$ = $1\,\text{cm}$, field: field lines shown. Kept constant: voltage $U$, charge of the ball $q$, horizontal position $x$, height above the lower plate $h$. Varied: plate distance $d$.
| plate distance $d$ | $F$ $(\mu\text{N})$ |
|---|
From the rows of your table, calculate the electric field strength $E = F : q$ and multiply it by the plate distance $d$ in metres.
How large is the product $Ed$? Answer in volts.
Hint 1: a guiding question
First calculate the field of a row: for example $F = 200\,\mu\text{N}$ and $q = 2.0\,\text{nC}$ give $E = 1.0 \cdot 10^{5}\,\text{N/C}$.
Hint 2: what is the connection
Convert the distance to metres: $2.0\,\text{cm} = 0.020\,\text{m}$.
Hint 3: an intermediate step
$Ed = 1.0 \cdot 10^{5}\,\text{N/C} \cdot 0.020\,\text{m}$. Compare the result with the voltage of the plates.
Show the solution
Given
readings of the model with the voltage $U = 2.0\,\text{kV}$ and the charge $q = 2.0\,\text{nC}$: at the distance $d = 2.0\,\text{cm}$ the force $F = 200\,\mu\text{N}$ and at the distance $d = 6.0\,\text{cm}$ the force $F = 66.7\,\mu\text{N}$
Find
the product $Ed$ in volts
Principle
At each distance the electric field strength is obtained from the force and the charge, $E = F : q$. If the electric field strength is $E = U : d$, the product $Ed$ is the same in every row and equal to the voltage of the plates.
Formula
$E = \dfrac{F}{q}$, so $Ed = \dfrac{Fd}{q}$
Conversions
$F = 200 \cdot 10^{-6}\,\text{N}$, $q = 2.0 \cdot 10^{-9}\,\text{C}$ and $d = 2.0\,\text{cm} = 0.020\,\text{m}$.
Substitution
Answer
$2.0 \cdot 10^{3}\,\text{V}$
Or $2000\,\text{V}$, two significant figures like the charge and the distance.
Reasonableness
The row $d = 6.0\,\text{cm}$ gives the same:
The product is the voltage of the plates $2.0\,\text{kV}$, so $E = U : d$. Units:
Optional Experiment 4: past the end of the plates
The voltage $2.0\,\text{kV}$, the plate distance $4.0\,\text{cm}$, the charge of the ball $+2.0\,\text{nC}$ and the height $h = 2.0\,\text{cm}$ stay constant; move the ball horizontally from the middle past the end of the plates, which is at $x = 10\,\text{cm}$. Record the force, for example at $0$, $4$, $8$, $10$, $11$ and $12\,\text{cm}$. How far from the middle of the plates is the field uniform?
Initial state: voltage $U$ = $2\,\text{kV}$, plate distance $d$ = $4\,\text{cm}$, charge of the ball $q$ = $2\,\text{nC}$, horizontal position $x$ = $0\,\text{cm}$, height above the lower plate $h$ = $2\,\text{cm}$, field: field lines shown. Kept constant: voltage $U$, plate distance $d$, charge of the ball $q$, height above the lower plate $h$. Varied: horizontal position $x$.
| horizontal position $x$ | $F$ $(\mu\text{N})$ |
|---|
Conclusion: the field between the plates
Write on the basis of your tables: does the force depend on the position of the ball between the plates? What does the quotient $F : q$ tell, and how does it depend on the distance of the plates when the voltage stays the same?
Model answer
The force was $100\,\mu\text{N}$ at every height from $0.5\,\text{cm}$ to $3.5\,\text{cm}$ and always pointed down, so the field between the plates is uniform: the same strength and direction everywhere between the plates. When the charge doubled or tripled, the force grew in the same proportion, and for a negative charge the force pointed up. The quotient $F : q = 5.0 \cdot 10^{4}\,\text{N/C}$ was the same for all charges: it is the electric field strength, which does not depend on the test charge. When the distance doubled from $2.0\,\text{cm}$ to $4.0\,\text{cm}$, the force halved from $200\,\mu\text{N}$ to $100\,\mu\text{N}$, and the product $Ed$ was $2000\,\text{V}$ in every row, the voltage of the plates. So $E = U : d$. The values are calculated by the model, and the model holds between the plates far from the ends of the plates.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
A charge changes the space around it
In the chapter Electric charge and Coulomb's law two charges acted on each other with a force although they did not touch each other. This action at a distance is described with the electric field. A charge changes the space around it so that an electric force acts on every other charge there. The electric field exists at every point, even if there is no charge at that point right now for the field to act on. When a charge is brought to the point, the field acts on it. The idea is the same as in gravity: the Earth pulls an apple through a field, although the apple does not touch the Earth.
Electric field strength
The field is studied with a small positive test charge $q$, so small that it does not change the positions of the charges that cause the field. When a force $\overline{F}$ acts on the test charge, the electric field strength at that point is the force divided by the test charge: $\overline{E} = \overline{F} : q$. So the electric field strength tells how large a force acts on one coulomb, and its unit is $\text{N/C}$. If a force of $100\,\mu\text{N}$ acts on a charge of $2.0\,\text{nC}$, the field strength is $5.0 \cdot 10^{4}\,\text{N/C}$. The quantity does not depend on the test charge: a doubled charge gets a doubled force, but the quotient stays the same. The symbol $E$ is the same as for the source voltage in the chapter Electromotive force and internal resistance, so in this chapter $E$ always means the electric field strength.
The electric field is a vector. Its direction is the direction in which the field pushes a positive charge. A force against the field acts on a negative charge, for example an electron. In the picture the field at the point $P$ is the same in both situations, but the direction of the force depends on the sign of the test charge.
Field lines
The field is illustrated with field lines. The direction of a field line at each point is the direction of the field, that is, the arrow shows in which direction the force on a positive charge points. Field lines start from positive charges and end on negative charges. Where the lines are dense, the field is strong, and where they are sparse, the field is weak. Two field lines never cross each other, because the field has only one direction at every point. A field line is not the path of a charge: it shows the direction of the force, not the direction in which the charge moves.
The field lines of a point charge are radial straight lines, and they thin out farther from the charge, because the field weakens as the distance grows, as Coulomb's law says. The field lines of two charges of opposite signs curve from the positive charge to the negative one.
A uniform field between plates
When two large metal plates are placed parallel to each other, facing each other, and connected to the terminals of a voltage source, positive charge collects on one plate and an equal amount of negative charge on the other. Between the plates the field lines are straight, parallel and evenly spaced. Then the strength and direction of the field are the same everywhere between the plates: the field is uniform. The same force acts on a small charged ball whether it is near the positive plate, near the negative plate or halfway between them.
Why does the force not grow when the ball is moved closer to a plate? Near the plate the nearest charges act strongly on the ball, but then most of the charges of the plate are at the sides of the ball. Their forces point almost along the plate, so the sideways parts cancel each other and do not pull the ball towards the plate. Farther from the plate each charge acts more weakly, but the force of more and more charges points almost perpendicular to the plate. For a large plate these two effects balance each other, so the field of the plate does not depend on the distance from the plate. Outside the plates the fields of the positive and the negative plate point in opposite directions and almost cancel each other, but between the plates they point in the same direction and add up. The field is uniform only between the plates and far from the ends of the plates. At the ends of the plates the field lines curve outwards and thin out.
The air purifier
In an electric air purifier the air first passes through a charging section, where the dust particles get an electric charge. Then the air flows between parallel plates, and there is a voltage of several kilovolts between the plates. The uniform field pushes every charged particle towards a plate with the same force the whole way, so the particle drifts to the surface of the plate and stays there, although the air goes on. The same principle is used in the electrostatic precipitators of power plants, which remove almost all the dust from flue gases.
Electric field strength
The electric field strength at a point where an electric force $\overline{F}$ acts on a small test charge $q$ is
$$\overline{E} = \frac{\overline{F}}{q}$$
where $\overline{E}$ is the electric field strength ($\text{N/C}$), $\overline{F}$ the electric force ($\text{N}$) and $q$ the test charge ($\text{C}$). When the field is known, the force on any charge is obtained by multiplying: $\overline{F} = q\overline{E}$. If $q$ is negative, the force points against the field.
The field from the voltage of the plates
The strength of a uniform field can be calculated from the voltage between the plates. When a positive charge $q$ moves from the positive plate to the negative one, the same force $F = qE$ acts on it the whole way, and the field does the work $W = Fd = qEd$. According to the chapter Electric circuit, current and voltage, voltage tells the energy per unit charge, so the same work is also $W = qU$. When these are set equal, the charge cancels, and the strength of a uniform field is
$$E = \frac{U}{d}$$
where $E$ is the electric field strength ($\text{V/m}$), $U$ the voltage between the plates ($\text{V}$) and $d$ the distance between the plates ($\text{m}$). The units $\text{V/m}$ and $\text{N/C}$ are the same unit, because $1\,\text{V/m} = 1\,\tfrac{\text{J}}{\text{C} \cdot \text{m}} = 1\,\tfrac{\text{N} \cdot \text{m}}{\text{C} \cdot \text{m}} = 1\,\text{N/C}$. For example, a voltage of $2.0\,\text{kV}$ across a plate distance of $4.0\,\text{cm}$ gives the field $2000\,\text{V} : 0.040\,\text{m} = 5.0 \cdot 10^{4}\,\text{V/m}$.
The equation holds only in a uniform field, that is, between the plates when the plates are large compared with the distance between them. The graph shows the field calculated with the plate model of the simulation at the height halfway between the plates. From the middle of the plates to about $7\,\text{cm}$ away the field is $U : d = 50\,\text{kV/m}$, but at the ends of the plates and outside them it weakens. In an air purifier a voltage of $5.0\,\text{kV}$ and a plate gap of $1.0\,\text{cm}$ give the field $5.0 \cdot 10^{5}\,\text{V/m}$. On the surface of the Earth in clear weather there is a field of about $100\,\text{V/m}$, and under a thundercloud the field can be as much as tens of kilovolts per metre.
A charged particle in a uniform field
In a uniform field a constant force $F = qE$ acts on a charged particle, so by Newton's second law its acceleration $a = qE : m$ is constant in vacuum when other forces are small. A particle at rest accelerates uniformly along the field, a positive one in the direction of the field and a negative one against the field. If the particle enters between the plates from the side, its path curves towards one plate in the same way as the path of a ball thrown horizontally curves downwards. In a field of $5.0 \cdot 10^{4}\,\text{V/m}$ a force of $8.0 \cdot 10^{-15}\,\text{N}$ acts on an electron, but its weight is only $8.9 \cdot 10^{-30}\,\text{N}$. The weight of an electron can therefore be ignored. For heavier bodies, such as oil droplets and dust particles, gravity can be as important as the electric force.
Common misconception
Many people think that the force on a charged ball between plates grows when the ball is moved closer to a plate, because by Coulomb's law the force grows when the distance gets smaller. Coulomb's law applies to point charges. The field of large plates is uniform: between the plates the force is the same everywhere and does not depend on how close to a plate the ball is. The strength depends only on the voltage and the distance of the plates, $E = U : d$.
Another misconception is that a field line is the path a charge travels, or that the field exists only where the lines are. A field line shows the direction of the force at each point, and there is field also between the lines. A third mistake is to think that the electric field strength depends on the test charge. A larger charge gets a larger force, but the quotient of the force and the charge, $E = F : q$, is the same.
Summary
- A charge changes the space around it: the electric field acts with an electric force on every charge there.
- The electric field strength is the force on a test charge divided by the charge, $\overline{E} = \overline{F} : q$, unit $\text{N/C} = \text{V/m}$. The force on a charge is $\overline{F} = q\overline{E}$: for a positive charge in the direction of the field, for a negative charge against the field.
- Field lines start from positive charges and end on negative charges, do not cross each other, and their density tells the strength of the field.
- Between parallel plates the field is uniform: the same strength and direction everywhere between the plates, when one is far from the ends of the plates.
- The strength of a uniform field is $E = U : d$. A constant force acts on a charged particle in it, so its acceleration is constant in vacuum when other forces are small.
Optional Return to your predictions
Compare your predictions with the values of the model in the experiments. Then return to the question at the start: why do the charged dust particles drift to the plates of the air purifier, and does the force depend on the position of the particle between the plates?
Would you change your prediction?
Apply
Worked example
Problem
A small oil droplet with a mass of $4.9 \cdot 10^{-15}\,\text{kg}$ hovers at rest between two horizontal plates. The distance of the plates is $6.0\,\text{mm}$, the voltage is $600\,\text{V}$, and the positive plate is on top. How large is the charge of the droplet, and how many elementary charges $e = 1.602 \cdot 10^{-19}\,\text{C}$ is it?
Step 1/7
Given and find. Mass $m = 4.9 \cdot 10^{-15}\,\text{kg}$, plate distance $d = 6.0\,\text{mm}$, voltage $U = 600\,\text{V}$, acceleration of free fall $g = 9.81\,\text{m/s}^2$ and elementary charge $e$. We are asked for the magnitude $q$ of the charge of the droplet and the number $n$ of elementary charges.
Step 2/7
Principle. The field between the plates is uniform, so $E = U : d$. The droplet is at rest, so by Newton's first law the forces on it cancel each other. The weight $mg$ points down, so the electric force $qE$ must point up, towards the positive plate: the droplet is negative. The charge is a multiple of the elementary charge, so the number is obtained by dividing.
Step 3/7
Formula solved.
$$\begin{aligned} E &= \frac{U}{d} \\ qE &= mg \\ q &= \frac{mg}{E} \\ n &= \frac{q}{e} \end{aligned}$$
Step 4/7
Conversions.
Step 5/7
Substitution. $$\begin{aligned} E &= \frac{U}{d} \\ &= \frac{600\,\text{V}}{6.0 \cdot 10^{-3}\,\text{m}} \\ &= 1.0 \cdot 10^{5}\,\text{V/m} \\ mg &= 4.9 \cdot 10^{-15}\,\text{kg} \\ &\quad \cdot 9.81\,\text{m/s}^2 \\ &= 4.807 \cdot 10^{-14}\,\text{N} \\ q &= \frac{mg}{E} \\ &= \frac{4.807 \cdot 10^{-14}\,\text{N}}{1.0 \cdot 10^{5}\,\text{V/m}} \\ &= 4.807 \cdot 10^{-19}\,\text{C} \\ n &= \frac{q}{e} \\ &= \frac{4.807 \cdot 10^{-19}\,\text{C}}{1.602 \cdot 10^{-19}\,\text{C}} \\ &= 3.0 \end{aligned}$$
Step 6/7
Answer. The charge of the droplet is $-4.8 \cdot 10^{-19}\,\text{C}$, that is, three elementary charges: the droplet has three extra electrons. The charge is given with two significant figures like the mass and the distance.
Step 7/7
Reasonableness. The electric force of three elementary charges in the field $1.0 \cdot 10^{5}\,\text{V/m}$ is
equal to the weight of the droplet
The number is almost exactly a whole number, as it should be, because charge is made of elementary charges.
Find the error
Apply to a new situation
In an inkjet printer a charged ink droplet passes between two deflection plates, and the field of the plates bends its path towards the right place on the paper. The mass of the droplet is $1.2 \cdot 10^{-10}\,\text{kg}$ and the magnitude of its charge is $2.0 \cdot 10^{-13}\,\text{C}$. The voltage between the plates is $1.6\,\text{kV}$ and the distance of the plates is $2.0\,\text{mm}$.
How large is the acceleration of the droplet in the direction perpendicular to the plates? Air resistance and weight are ignored. Answer in the unit $\text{m/s}^2$.
Hint 1: a guiding question
Which force accelerates the droplet towards a plate, and how does it depend on the field of the plates?
Hint 2: what is the connection
The electric field strength between the plates is $E = U : d$, the force is $F = qE$, and by Newton's second law $a = F : m$.
Hint 3: an intermediate step
The field is $1600\,\text{V} : 0.0020\,\text{m} = 8.0 \cdot 10^{5}\,\text{V/m}$.
Show the solution
Given
mass of the droplet $m = 1.2 \cdot 10^{-10}\,\text{kg}$, magnitude of the charge $q = 2.0 \cdot 10^{-13}\,\text{C}$, voltage $U = 1.6\,\text{kV}$, plate distance $d = 2.0\,\text{mm}$; air resistance and weight are ignored
Find
the acceleration $a$ of the droplet in the unit $\text{m/s}^2$
Principle
Between the plates the field is uniform, so a constant force $qE$ acts on the droplet in the direction perpendicular to the plates. When the other forces are ignored, by Newton's second law this force determines the acceleration, and it is the same over the whole distance between the plates.
Formula
$E = \dfrac{U}{d}$, $F = qE$ and $F = ma$, so $a = \dfrac{qU}{md}$
Conversions
Substitution
Answer
$1.3 \cdot 10^{3}\,\text{m/s}^2$
Two significant figures like the given values.
Reasonableness
The weight of the droplet is only
less than one per cent of the electric force, so ignoring the weight is justified. The acceleration is about $140$ times the acceleration of free fall, so the path of the droplet bends clearly even over a short distance.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The voltage between two parallel plates is $9.0\,\text{kV}$, and the distance of the plates is $3.0\,\text{cm}$. How large is the electric field strength between the plates?
Answer
The field between the plates is uniform, so
$E$$= U : d$$= 9000\,\text{V} : 0.030\,\text{m}$$= 3.0 \cdot 10^{5}\,\text{V/m}$The same value holds everywhere between the plates when one is far from the ends of the plates.
- Why can two field lines not cross each other?
Answer
The direction of a field line at each point is the direction of the electric field. If two lines crossed, the field would have two different directions at the crossing point, and two forces in different directions would act at the same time on a charge brought there. But the field has only one direction at every point.
1. Concept
2. Read the field line picture
Optional 3. Explain the air purifier
Using the electric field, explain why the charged dust particles of the air purifier end up on the surface of the plates although the air flows past the plates.
Why are the plates of the purifier close to each other?
Model answer
There is a voltage between the plates, so one plate is positive and the other negative, and between them there is a uniform electric field. A force $\overline{F} = q\overline{E}$ acts on a charged dust particle, perpendicular to the plates and equally large everywhere between the plates. The air carries the particle along the plates, but the electric force pushes it towards one plate the whole way, a positive particle towards the negative plate and a negative one towards the positive plate. The particle has time to drift to the surface of the plate before the air carries it past the plates, and it sticks to the plate. The plates are close to each other, because the electric field strength is $E = U : d$: a small distance gives a strong field with the same voltage, and the distance the particle has to travel to the plate is short.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 11-1Core
An electric force of $1.2 \cdot 10^{-4}\,\text{N}$ acts in an electric field on a small ball whose charge is $3.0 \cdot 10^{-9}\,\text{C}$. How large is the electric field strength at the ball?
Answer
$4.0 \cdot 10^{4}\,\text{N/C}$
Show the solution
Given
charge of the ball $q = 3.0 \cdot 10^{-9}\,\text{C}$, electric force $F = 1.2 \cdot 10^{-4}\,\text{N}$
Find
the electric field strength $E$ in the unit $\text{N/C}$
Principle
The electric field strength is the force on a charge brought to the point, divided by that charge. The ball is small, so it is assumed to measure the field without changing it.
Formula
$F = qE$, so $E = \dfrac{F}{q}$
Conversions
Not needed: the force is in newtons and the charge in coulombs.
Substitution
$E$$= \dfrac{1.2 \cdot 10^{-4}\,\text{N}}{3.0 \cdot 10^{-9}\,\text{C}}$$= 4.0 \cdot 10^{4}\,\text{N/C}$Answer
$4.0 \cdot 10^{4}\,\text{N/C}$
Two significant figures like the force and the charge.
Reasonableness
Check by multiplying:
$3.0 \cdot 10^{-9}\,\text{C} \cdot 4.0 \cdot 10^{4}\,\text{N/C}$$= 1.2 \cdot 10^{-4}\,\text{N}$which is the given force. The field is of the same order of magnitude as a voltage of a few kilovolts across a plate gap of a few centimetres, which is a typical field in a school experiment.
- Exercise 11-2Core
The strength of a uniform electric field is $2.0 \cdot 10^{4}\,\text{N/C}$. A small ball whose charge is $5.0\,\text{nC}$ is brought into the field.
How large is the electric force on the ball? Answer in micronewtons.
Answer
$1.0 \cdot 10^{2}\,\mu\text{N}$
Show the solution
Given
electric field strength $E = 2.0 \cdot 10^{4}\,\text{N/C}$, charge of the ball $q = 5.0\,\text{nC}$
Find
the electric force $F$ on the ball in micronewtons
Principle
The electric field strength tells the force per coulomb, so the force on a charge is the charge times the electric field strength. The field is uniform, so the force is the same everywhere in the field.
Formula
$F = qE$
Conversions
$q$$= 5.0\,\text{nC}$$= 5.0 \cdot 10^{-9}\,\text{C}$finally $1\,\text{N} = 10^{6}\,\mu\text{N}$.
Substitution
$F$$= 5.0 \cdot 10^{-9}\,\text{C} \cdot 2.0 \cdot 10^{4}\,\text{N/C}$$= 1.0 \cdot 10^{-4}\,\text{N}$$= 100\,\mu\text{N}$Answer
$1.0 \cdot 10^{2}\,\mu\text{N}$
Or $100\,\mu\text{N}$, two significant figures like the given values.
Reasonableness
The force is about $0.1\,\text{mN}$, which corresponds to the weight of a body of about $10\,\text{mg}$. It is small, but a light ball moves clearly under its effect, as is seen in electrostatic experiments.
- Exercise 11-3
The voltage between two parallel metal plates is $600\,\text{V}$, and the distance of the plates is $0.040\,\text{m}$. How large is the electric field strength between the plates?
Answer
$1.5 \cdot 10^{4}\,\text{V/m}$
Show the solution
Given
voltage between the plates $U = 600\,\text{V}$, plate distance $d = 0.040\,\text{m}$
Find
the electric field strength $E$ in the unit $\text{V/m}$
Principle
Between parallel plates the field is uniform. When a charge moves from one plate to the other, the field does the work $qEd$, and by the definition of voltage the same work is $qU$, so the electric field strength is the voltage divided by the plate distance.
Formula
$qEd = qU$, so $E = \dfrac{U}{d}$
Conversions
Not needed: the distance is in metres.
Substitution
$E$$= \dfrac{600\,\text{V}}{0.040\,\text{m}}$$= 15\,000\,\text{V/m}$Answer
$1.5 \cdot 10^{4}\,\text{V/m}$
Two significant figures like the distance $0.040\,\text{m}$.
Reasonableness
The voltage is divided evenly over the plate gap: for every centimetre there is $600\,\text{V} : 4 = 150\,\text{V}$, and for every metre $15\,000\,\text{V}$. The unit $\text{V/m}$ is the same as $\text{N/C}$.
- Exercise 11-4
The inner and outer surfaces of a cell membrane have charges of opposite signs, and inside the membrane there is an approximately uniform electric field whose strength is $1.0 \cdot 10^{7}\,\text{V/m}$. The thickness of the membrane is $7.0 \cdot 10^{-9}\,\text{m}$.
How large is the voltage across the membrane? Answer in millivolts.
Answer
$70\,\text{mV}$
Show the solution
Given
electric field strength $E = 1.0 \cdot 10^{7}\,\text{V/m}$, thickness of the membrane $d = 7.0 \cdot 10^{-9}\,\text{m}$
Find
the voltage $U$ across the membrane in millivolts
Principle
The surfaces of the membrane have charges of opposite signs like the plates, and inside the membrane the field is approximately uniform. Then the relation $E = U : d$ between the electric field strength and the voltage holds, and the voltage is obtained by solving it.
Formula
$E = \dfrac{U}{d}$, so $U = Ed$
Conversions
Finally $1\,\text{V} = 1000\,\text{mV}$.
Substitution
$U$$= 1.0 \cdot 10^{7}\,\text{V/m} \cdot 7.0 \cdot 10^{-9}\,\text{m}$$= 0.070\,\text{V}$$= 70\,\text{mV}$Answer
$70\,\text{mV}$
Two significant figures like the given values.
Reasonableness
The voltage is small because the membrane is very thin, although the field is strong: $10^{7}\,\text{V/m}$ means only $0.01\,\text{V}$ per nanometre. The resting voltage of a nerve cell is indeed about $70\,\text{mV}$.
- Exercise 11-5
A small ball is held always at the same place between two plates, and its charge is changed. The graph shows the electric force on the ball as a function of the charge, with a straight line fitted to the points.
How large is the electric field strength at the ball? Determine it from the graph.
Answer
$3.0 \cdot 10^{4}\,\text{N/C}$
Show the solution
Given
the measurement points of the graph and a straight line fitted to them through the origin; the ball is at the same place in every measurement
Find
the electric field strength $E$ in the unit $\text{N/C}$
Principle
At the same place the field is the same, so the force is directly proportional to the charge, $F = qE$. The $(q, F)$ graph is a straight line through the origin, and its slope is the electric field strength. The slope is read from the fitted line, not from a single measurement point.
Reading the graph
The line goes from the origin to the crossing of grid lines $(4.0\,\text{nC}$; $120\,\mu\text{N})$. So the charge grows by $4.0\,\text{nC}$ and the force by $120\,\mu\text{N}$.
Formula
$E = \dfrac{\Delta F}{\Delta q}$
Conversions
$120\,\mu\text{N} = 120 \cdot 10^{-6}\,\text{N}$ and $4.0\,\text{nC} = 4.0 \cdot 10^{-9}\,\text{C}$.
Substitution
$E$$= \dfrac{120 \cdot 10^{-6}\,\text{N}}{4.0 \cdot 10^{-9}\,\text{C}}$$= 3.0 \cdot 10^{4}\,\text{N/C}$Answer
$3.0 \cdot 10^{4}\,\text{N/C}$
Two significant figures, because a crossing of the grid is read from the graph, where the force is $120\,\mu\text{N}$ with two significant figures.
Reasonableness
The measurement point $(3.0\,\text{nC}$; $91\,\mu\text{N})$ gives almost the same value: $91\,\mu\text{N} : 3.0\,\text{nC} \approx 3.0 \cdot 10^{4}\,\text{N/C}$. The line was fitted through the origin because, by the model $F = qE$, an uncharged ball feels no electric force in a uniform field, although the field exists.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 11-6Core
The voltage between two parallel plates is $4.5\,\text{kV}$, and the distance of the plates is $9.0\,\text{mm}$. A small ball whose charge is $4.0\,\text{nC}$ is brought between the plates.
How large is the electric force on the ball? Answer in millinewtons.
Answer
$2.0\,\text{mN}$
Show the solution
Given
voltage between the plates $U = 4.5\,\text{kV}$, plate distance $d = 9.0\,\text{mm}$, charge of the ball $q = 4.0\,\text{nC}$
Find
the electric force $F$ on the ball in millinewtons
Principle
Between the plates the field is uniform, so its strength is obtained from the voltage and the distance. The force on a charge is the charge times the electric field strength, and it does not depend on where between the plates the ball is.
Formula
$E = \dfrac{U}{d}$ and $F = qE$, so $F = \dfrac{qU}{d}$
Conversions
$U$$= 4.5\,\text{kV}$$= 4.5 \cdot 1000\,\text{V}$$= 4500\,\text{V}$$d$$= 9.0\,\text{mm}$$= 9.0 \cdot 10^{-3}\,\text{m}$$q = 4.0 \cdot 10^{-9}\,\text{C}$; finally $1\,\text{N} = 1000\,\text{mN}$.
Substitution
$E$$= \dfrac{4500\,\text{V}}{9.0 \cdot 10^{-3}\,\text{m}}$$= 5.0 \cdot 10^{5}\,\text{V/m}$$F$$= 4.0 \cdot 10^{-9}\,\text{C} \cdot 5.0 \cdot 10^{5}\,\text{V/m}$$= 2.0 \cdot 10^{-3}\,\text{N}$$= 2.0\,\text{mN}$Answer
$2.0\,\text{mN}$
Two significant figures like the given values.
Reasonableness
Units:
$\text{C} \cdot \text{V/m}$$= \text{J/m}$$= \text{N}$The field $5 \cdot 10^{5}\,\text{V/m}$ is strong but clearly smaller than the roughly $3 \cdot 10^{6}\,\text{V/m}$ needed for the breakdown of air, so the arrangement is possible.
- Exercise 11-7Core
The distance of two parallel plates is $5.0\,\text{cm}$. An electric force of $2.4\,\text{mN}$ acts on a small ball between the plates, with a charge of $8.0\,\text{nC}$.
How large is the voltage between the plates? Answer in kilovolts.
Answer
$15\,\text{kV}$
Show the solution
Given
plate distance $d = 5.0\,\text{cm}$, charge of the ball $q = 8.0\,\text{nC}$, electric force $F = 2.4\,\text{mN}$
Find
the voltage $U$ between the plates in kilovolts
Principle
The electric field strength is obtained from the force and the charge, and in the uniform field between the plates the voltage is the electric field strength times the plate distance.
Formula
$E = \dfrac{F}{q}$ and $E = \dfrac{U}{d}$, so $U = \dfrac{Fd}{q}$
Conversions
$F$$= 2.4\,\text{mN}$$= 2.4 \cdot 10^{-3}\,\text{N}$$d = 5.0\,\text{cm} = 0.050\,\text{m}$, $q = 8.0 \cdot 10^{-9}\,\text{C}$; finally $1\,\text{kV} = 1000\,\text{V}$.
Substitution
$E$$= \dfrac{2.4 \cdot 10^{-3}\,\text{N}}{8.0 \cdot 10^{-9}\,\text{C}}$$= 3.0 \cdot 10^{5}\,\text{N/C}$$U$$= Ed$$= 3.0 \cdot 10^{5}\,\text{N/C} \cdot 0.050\,\text{m}$$= 15\,000\,\text{V}$$= 15\,\text{kV}$Answer
$15\,\text{kV}$
Two significant figures like the given values.
Reasonableness
Check by multiplying:
$8.0 \cdot 10^{-9}\,\text{C} \cdot 15\,000\,\text{V} : 0.050\,\text{m}$$= 2.4 \cdot 10^{-3}\,\text{N}$which is the given force. Units:
$\text{N} \cdot \text{m} : \text{C}$$= \text{J/C}$$= \text{V}$ - Exercise 11-8
The voltage between the plates of an air purifier is $3.0\,\text{kV}$, and the distance of the plates is $6.0\,\text{mm}$. A dust particle coming between the plates has a charge of $1.6 \cdot 10^{-17}\,\text{C}$ and a mass of $4.0 \cdot 10^{-14}\,\text{kg}$.
How many times larger is the electric force on the particle than its weight?
Answer
$20$
Show the solution
Given
voltage $U = 3.0\,\text{kV}$, plate distance $d = 6.0\,\text{mm}$, charge of the particle $q = 1.6 \cdot 10^{-17}\,\text{C}$ and mass $m = 4.0 \cdot 10^{-14}\,\text{kg}$, $g = 9.81\,\text{m/s}^2$
Find
the ratio of the electric force to the weight $F : G$
Principle
Between the plates the field is uniform, $E = U : d$, and the electric force $qE$ acts on the particle. Gravity is $mg$. The ratio tells which force determines the motion of the particle.
Formula
$\dfrac{F}{G}$$= \dfrac{qE}{mg}$$= \dfrac{qU}{mgd}$Conversions
$U$$= 3.0 \cdot 1000\,\text{V}$$= 3000\,\text{V}$$d = 6.0 \cdot 10^{-3}\,\text{m}$.
Substitution
$F$$= \dfrac{1.6 \cdot 10^{-17}\,\text{C} \cdot 3000\,\text{V}}{6.0 \cdot 10^{-3}\,\text{m}}$$= 8.0 \cdot 10^{-12}\,\text{N}$$G$$= 4.0 \cdot 10^{-14}\,\text{kg} \cdot 9.81\,\text{m/s}^2$$= 3.924 \cdot 10^{-13}\,\text{N}$$\dfrac{F}{G}$$= \dfrac{8.0 \cdot 10^{-12}\,\text{N}}{3.924 \cdot 10^{-13}\,\text{N}}$$= 20.4$Answer
about $20$ times, two significant figures like the given values.
Reasonableness
The ratio has no unit, because both are forces. The electric force is much larger than the weight, so the particle drifts towards a plate and does not just fall, which is a condition for the purifier to work.
- Exercise 11-9
An electric force of $0.15\,\text{mN}$ acts on a charged ball between two plates when the distance of the plates is $6.0\,\text{cm}$. The plates are moved closer to each other so that the distance is $2.0\,\text{cm}$, and the voltage is kept the same. The ball stays between the plates, far from the ends of the plates.
How large is the force on the ball now? Answer in millinewtons.
Answer
$0.45\,\text{mN}$
Show the solution
Given
at first the force $F_1 = 0.15\,\text{mN}$ and the distance $d_1 = 6.0\,\text{cm}$, at the end the distance $d_2 = 2.0\,\text{cm}$; the voltage and the charge of the ball stay the same
Find
the new force $F_2$ in millinewtons
Principle
The ball is far from the ends of the plates, so the field is uniform and $E = U : d$. The force is $F = qE = qU : d$, and when $q$ and $U$ stay the same, the force is inversely proportional to the plate distance.
Formula
$\dfrac{F_2}{F_1}$$= \dfrac{qU : d_2}{qU : d_1}$$= \dfrac{d_1}{d_2}$so $F_2 = F_1 \cdot \dfrac{d_1}{d_2}$
Conversions
Not needed: the distances are in the same unit, so the unit cancels.
Substitution
$F_2$$= 0.15\,\text{mN} \cdot \dfrac{6.0\,\text{cm}}{2.0\,\text{cm}}$$= 0.15\,\text{mN} \cdot 3.0$$= 0.45\,\text{mN}$Answer
$0.45\,\text{mN}$
Two significant figures like the given values.
Reasonableness
A distance reduced to one third triples the field, because the same voltage is spread over one third of the distance. The force does not grow ninefold, as it would by Coulomb's law for point charges, because the field of plates does not depend on the distance from a plate.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 11-10
A negative oil droplet hovers at rest between two horizontal plates when the voltage between the plates is $450\,\text{V}$ and the positive plate is on top. The charge of the droplet is three elementary charges.
The droplet gets one extra electron, so that its charge is four elementary charges, but its mass does not change.
How large a voltage is needed for the droplet to hover at rest again?
Answer
$3.4 \cdot 10^{2}\,\text{V}$
Show the solution
Given
at first the voltage $U_1 = 450\,\text{V}$ and the charge $q_1 = 3e$, at the end the charge $q_2 = 4e$; the mass and the plate distance stay the same
Find
the voltage $U_2$ at which the droplet hovers
Principle
A hovering droplet is at rest, so the electric force up and the weight down are equal: $qE = mg$. The field of the plates is uniform, $E = U : d$. When $m$, $g$ and $d$ stay the same, the product $qU$ must stay the same: a larger charge needs a smaller voltage.
Formula
$\dfrac{q_1 U_1}{d}$$= mg$$= \dfrac{q_2 U_2}{d}$so $U_2 = U_1 \cdot \dfrac{q_1}{q_2}$
Conversions
Not needed: the elementary charge $e$ cancels.
Substitution
$U_2$$= 450\,\text{V} \cdot \dfrac{3e}{4e}$$= 450\,\text{V} \cdot 0.75$$= 337.5\,\text{V}$Answer
$3.4 \cdot 10^{2}\,\text{V}$
($337.5\,\text{V}$), rounded to two significant figures according to the precision of the voltage $450\,\text{V}$; the numbers of elementary charges are exact.
Reasonableness
The charge grows by one third, so the same electric force is obtained with a weaker field, and the voltage decreases by one quarter. If the voltage were kept at $450\,\text{V}$, the electric force would be $\tfrac{4}{3}$ of the weight, and the droplet would start to rise towards the positive plate.
- Exercise 11-11
An electron starts from rest at the negative plate and accelerates in a uniform field to the positive plate. The plates are in vacuum.
The voltage between the plates is $50\,\text{V}$, and the distance of the plates is $1.0\,\text{cm}$. The mass of the electron is $9.109 \cdot 10^{-31}\,\text{kg}$ and the magnitude of its charge is $1.602 \cdot 10^{-19}\,\text{C}$.
How large is the speed of the electron when it arrives at the positive plate? Answer in kilometres per second.
Answer
$4200\,\text{km/s}$
Show the solution
Given
voltage $U = 50\,\text{V}$, plate distance $d = 1.0\,\text{cm}$, mass of the electron $m = 9.109 \cdot 10^{-31}\,\text{kg}$ and charge $e = 1.602 \cdot 10^{-19}\,\text{C}$; initial speed $v_0 = 0$
Find
the speed $v$ of the electron at the positive plate in kilometres per second
Principle
In a uniform field a constant force $eE$ acts on the electron towards the positive plate, and the plates are in vacuum, so there are no other forces (the weight is negligibly small). By Newton's second law the acceleration $a = eE : m$ is constant, so the motion is uniformly accelerated, and for a body that has travelled the distance $d$ from rest $v^2 = 2ad$ holds.
Formula
$a = \dfrac{eE}{m} = \dfrac{eU}{md}$ and $v^2 = 2ad$, so
$v$$= \sqrt{2ad}$$= \sqrt{\dfrac{2eU}{m}}$Conversions
$d = 1.0\,\text{cm} = 0.010\,\text{m}$; finally $1\,\text{km/s} = 1000\,\text{m/s}$.
Substitution
$E$$= \dfrac{50\,\text{V}}{0.010\,\text{m}}$$= 5000\,\text{V/m}$$a$$= \dfrac{1.602 \cdot 10^{-19}\,\text{C} \cdot 5000\,\text{V/m}}{9.109 \cdot 10^{-31}\,\text{kg}}$$= 8.794 \cdot 10^{14}\,\text{m/s}^2$$v$$= \sqrt{2 \cdot 8.794 \cdot 10^{14}\,\text{m/s}^2 \cdot 0.010\,\text{m}}$$= 4.19 \cdot 10^{6}\,\text{m/s}$Answer
about $4200\,\text{km/s}$ or $4.2 \cdot 10^{6}\,\text{m/s}$, two significant figures like the voltage and the distance.
Reasonableness
The speed is about $1.4\,\%$ of the speed of light, so classical mechanics is enough. The final result $\sqrt{2eU : m}$ does not depend on the plate distance: the shorter distance is compensated by the stronger field.
- Exercise 11-12
A small charged plastic ball with a mass of $2.0\,\text{g}$ hangs on an insulating thread in a horizontal uniform electric field whose strength is $1.5 \cdot 10^{5}\,\text{N/C}$. The ball is at rest, and the thread has turned to an angle of $10^\circ$ from the vertical.
How large is the charge of the ball? Answer in nanocoulombs.
Answer
$23\,\text{nC}$
Show the solution
Given
mass of the ball $m = 2.0\,\text{g}$, electric field strength $E = 1.5 \cdot 10^{5}\,\text{N/C}$ horizontal, angle of the thread from the vertical $\alpha = 10^\circ$, $g = 9.81\,\text{m/s}^2$
Find
the charge $q$ of the ball in nanocoulombs
Principle
The forces on the ball are the weight $mg$ down, the electric force $qE$ horizontally and the tension of the thread along the thread. The ball is at rest, so the sum of the forces is zero. In the vertical direction the vertical component of the tension, $T\cos\alpha$, cancels the weight, and in the horizontal direction its horizontal component, $T\sin\alpha$, cancels the electric force.
Formula
$T\sin\alpha = qE$ and $T\cos\alpha = mg$; dividing gives $\tan\alpha = \dfrac{qE}{mg}$, so $q = \dfrac{mg\tan\alpha}{E}$
Conversions
$m$$= 2.0\,\text{g}$$= 2.0 \cdot 10^{-3}\,\text{kg}$finally $1\,\text{C} = 10^{9}\,\text{nC}$.
Substitution
$mg\tan\alpha$$= 2.0 \cdot 10^{-3}\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot \tan 10^\circ$$= 3.459 \cdot 10^{-3}\,\text{N}$$q$$= \dfrac{3.459 \cdot 10^{-3}\,\text{N}}{1.5 \cdot 10^{5}\,\text{N/C}}$$= 2.306 \cdot 10^{-8}\,\text{C}$$= 23\,\text{nC}$Answer
$23\,\text{nC}$
Two significant figures like the mass and the electric field strength.
Reasonableness
The electric force is
$mg\tan 10^\circ$$\approx 0.18 \cdot 0.020\,\text{N}$$= 3.5\,\text{mN}$about one fifth of the weight, which fits a small deflection. A charge of tens of nanocoulombs is typical for a plastic ball charged by rubbing.