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Educora
AdvancedGrades 10–1125 min30 / 44

Electric potential and potential difference

Learn the work of the electric field, the potential energy of a charge, potential and potential difference, the electronvolt, the relation E = U / d, equipotential surfaces, conductors and dielectrics in a field, and how charged particles are accelerated and deflected — with DİM-style tasks.

Check yourself
In this lesson you will learn
  • Calculate the work of a uniform field and of a point charge’s field, and explain why it does not depend on the path
  • Find potential energy, potential (φ = k · q / r, superposition) and potential difference (U = A / q), and express energy in electronvolts
  • Use E = U / d and equipotential surfaces, and explain how conductors and dielectrics behave in an electrostatic field
  • Calculate the speed, energy and deflection of a charged particle accelerated by a voltage and deflected between plates

A sparrow lands on a high-voltage power line and nothing happens to it, yet a person standing on the ground who touched the same wire would be in mortal danger. The word “voltage” explains the difference: the bird’s two feet are at almost the same potential, while a person’s hand and feet are not. In the lesson «Electric charge, Coulomb’s law and the electric field» we described the field through force — the field strength. Now we describe it through energy: the work of the field, potential and potential difference. This is the second half of the DİM test-collection chapter «Electric charge. Electric field»; besides calculations it often asks about φ(r) and E(r) graphs, equipotential surfaces and I–II–III statements.

Work done by the electric field

Place a positive charge q in the uniform field between the plates of a parallel-plate capacitor. A constant force F = q · E acts on it. When the charge moves a distance d along a field line, the field does work A = F · d = q · E · d. If the charge travels a path s at an angle α to the field lines, the work is A = F · s · cos α, and s · cos α is the projection of the displacement onto the direction of the field lines. So in a uniform field only the distance covered along the field lines counts — just as only the change in height counts in the work of gravity.

A = q · E · d, d = s · cos α
where:
  • Awork done by the field, J
  • qcharge moved (with its sign), C
  • Estrength of the uniform field, V/m
  • dprojection of the displacement onto the direction of the field lines, m
  • αangle between the displacement and the field lines

For a positive charge moving along the field lines A > 0, against them A < 0, perpendicular to them A = 0. For a negative charge (an electron) the signs are reversed.

Example 1: work of a uniform field

A charge q = 5 nC is moved in a uniform field of strength 2 · 10⁴ V/m. Find the work of the field if the charge moves:
1) 4 cm along the field lines;
2) 5 cm at 60° to the field lines;
3) 3 cm perpendicular to the field lines.
4) What work does the field do if an electron moves 2 cm in the direction of the field lines? (e = 1.6 · 10⁻¹⁹ C)

Show solution
1) A = q · E · d = 5 · 10⁻⁹ · 2 · 10⁴ · 0.04 = 4 · 10⁻⁶ J (4 µJ).
2) d = s · cos 60° = 5 · 0.5 = 2.5 cm; A = 5 · 10⁻⁹ · 2 · 10⁴ · 0.025 = 2.5 · 10⁻⁶ J.
3) cos 90° = 0 → A = 0: the field does no work on a charge moving perpendicular to the field lines.
4) The electron is negative, so the force on it points against E: A = −e · E · d = −1.6 · 10⁻¹⁹ · 2 · 10⁴ · 0.02 = −6.4 · 10⁻¹⁷ J. The field resists this motion of the electron.

Now take the charge from point 1 to point 2 along a straight line, a broken line or a curve. Split the path into tiny steps: on steps perpendicular to the field lines the work is zero, and the steps along the lines always add up to the same projection d. So the work of the field does not depend on the shape of the path — only on the positions of the start and end points. The same holds for the field of a point charge: radial steps involve work, steps along circular arcs around the charge do not. An important consequence: when a charge goes round a closed path and returns to its starting point, the total work of the field is zero.

Definition
Conservative (potential) field

A field whose work does not depend on the shape of the path, only on the start and end points; the work round any closed path is zero. The electrostatic field and the field of gravity are both conservative. Only in such a field can we speak of the potential energy of a charge.

Potential energy and potential

Because the work of the field does not depend on the path, every position of a charge in the field has a definite potential energy Wp — like the energy m · g · h in the gravitational field. Positive work of the field reduces the potential energy: like a falling stone, a positive charge left free moves towards lower potential energy. The zero level of potential energy is chosen by convention; for point charges it is usually taken infinitely far away.

A = Wp₁ − Wp₂ = −ΔWp, Wp = k · q₁ · q₂ / (ε · r)A = Wp₁ − Wp₂ = −ΔWp, Wp = k · q₁ · q₂ / (ε · r)
where:
  • Wp₁, Wp₂potential energy of the charge at the start and at the end, J
  • q₁, q₂point charges (with their signs), C
  • k9 · 10⁹ N · m²/C²
  • εrelative permittivity of the medium (1 for air)
  • rdistance between the charges, m

When the field does positive work, the potential energy decreases. For like charges Wp > 0 (an external force must do work to bring them closer), for unlike charges Wp < 0. The zero level is at infinity.

Example 2: potential energy of two charges

Point charges of +2 nC and +3 nC are 6 cm apart in air.
1) Find the potential energy of the system.
2) The charges repel each other and move apart to 18 cm. How much work does the field do?
3) If the second charge were −3 nC, what would the potential energy be, and how much external work would be needed to separate the charges to infinity?

Show solution
1) Wp = k · q₁ · q₂ / r = 9 · 10⁹ · 2 · 10⁻⁹ · 3 · 10⁻⁹ / 0.06 = 9 · 10⁻⁷ J.
2) At 18 cm: Wp₂ = 9 · 10⁹ · 6 · 10⁻¹⁸ / 0.18 = 3 · 10⁻⁷ J. A = Wp₁ − Wp₂ = 9 · 10⁻⁷ − 3 · 10⁻⁷ = 6 · 10⁻⁷ J — positive: the repulsive force moves the charges in its own direction.
3) Wp = 9 · 10⁹ · 2 · 10⁻⁹ · (−3 · 10⁻⁹) / 0.06 = −9 · 10⁻⁷ J. At infinity Wp = 0, so the energy must rise by 9 · 10⁻⁷ J: an external force has to do 9 · 10⁻⁷ J of work.

The potential energy is proportional to the charge q₀ brought into the field: double the charge and the energy doubles. So the ratio Wp / q₀ no longer depends on that charge and describes only the given point of the field — this is the potential. Field strength is the force characteristic of a field, potential is its energy characteristic. Field strength is a vector; potential is a scalar: it has a sign but no direction.

Definition
Potential (φ)

A scalar quantity equal to the ratio of the potential energy of a charge placed at a given point of the field to that charge: φ = Wp / q₀. Its unit is the volt (V): 1 V = 1 J/C. Physical meaning: the work of the field per unit charge when a positive charge is moved from this point to the zero level (to infinity).

φ = Wp / q₀, φ = k · q / (ε · r)φ = Wp / q₀, φ = k · q / (ε · r)
where:
  • φpotential at the given point, V
  • q₀charge brought into the field, C
  • qpoint charge that creates the field (with its sign), C
  • rdistance from the charge to the point, m

φ = 0 at infinity. In the field of a positive charge the potential is positive and grows towards the charge; in the field of a negative charge it is negative. φ ~ 1/r while E ~ 1/r²: when the distance doubles, the potential halves and the field strength drops 4 times. At the same point |φ| = E · r.

Example 3: potential of a point charge

1) Find the potential at point A, 30 cm from a point charge q = 2 nC in air.
2) A charge q₀ = −3 nC is placed at A. What is its potential energy?
3) How do the potential and the field strength change if the distance is tripled?

Show solution
1) φ = k · q / r = 9 · 10⁹ · 2 · 10⁻⁹ / 0.3 = 18 / 0.3 = 60 V.
2) Wp = q₀ · φ = −3 · 10⁻⁹ · 60 = −1.8 · 10⁻⁷ J (a negative charge in the field of a positive one: the energy is negative, the charges attract).
3) φ ~ 1/r → 3 times smaller (20 V); E ~ 1/r² → 9 times smaller.
Check: at A, E = 9 · 10⁹ · 2 · 10⁻⁹ / 0.3² = 200 V/m and E · r = 200 · 0.3 = 60 V = φ.
Interactive
Loading simulation…
Field of a point charge: x is the distance r from the charge (m), q the charge (nC). y₁ = 9q / x is the potential φ (V), y₂ = 9q / x² the field-strength projection E (V/m). For q = 2 at r = 0.3 m, φ = 60 V and E = 200 V/m (Example 3). The curves cross at r = 1 m because |φ| / E = r. Make q negative: the potential becomes negative, and E < 0 means the field points towards the charge.

In the field of several charges each charge creates its own potential, and the resulting potential is their algebraic sum — added as numbers with signs, not as vectors. This is much easier than finding the field strength: no directions to draw.

φ = φ₁ + φ₂ + … = k · q₁ / r₁ + k · q₂ / r₂ + …φ = φ₁ + φ₂ + … = k · q₁ / r₁ + k · q₂ / r₂ + …
where:
  • φ₁, φ₂potential created by each charge alone (with its sign), V
  • r₁, r₂distance from the point to the corresponding charge, m

The potential of a negative charge is taken as negative. Where the potential is zero the field strength need not be zero, and vice versa — see Example 4.

Example 4 (multiple choice): potential and field at the midpoint

Point charges of +4 nC and −4 nC are 6 cm apart in air. What are the potential (φ) and the field strength (E) at the midpoint of the segment joining them?
A) φ = 0, E = 0
B) φ = 0, E = 8 · 10⁴ V/m
C) φ = 2400 V, E = 0
D) φ = 2400 V, E = 8 · 10⁴ V/m
E) φ = 1200 V, E = 4 · 10⁴ V/m

Show solution
1) The midpoint is 3 cm = 0.03 m from each charge.
Potentials: φ₁ = 9 · 10⁹ · 4 · 10⁻⁹ / 0.03 = 1200 V, φ₂ = −1200 V → φ = 1200 − 1200 = 0.
2) Field strengths: each is 9 · 10⁹ · 4 · 10⁻⁹ / 0.03² = 4 · 10⁴ V/m; both vectors point towards the negative charge (out of the positive one, into the negative one) and add up: E = 8 · 10⁴ V/m. Answer: B.
With like charges (+4 nC and +4 nC) the answer would be C: φ = 2400 V, E = 0. So E can be non-zero where φ = 0, and vice versa. Option A comes from treating potential and field strength the same way, option E from counting only one charge.

Potential difference, the electronvolt and equipotential surfaces

In practice what matters is not the potential itself but the difference between the potentials of two points, because the work of the field depends on it: A = Wp₁ − Wp₂ = q · φ₁ − q · φ₂ = q · (φ₁ − φ₂). The potential difference is also called voltage and is written U. In a circuit voltage is measured with a voltmeter; in electrostatics an electrometer is used: its case is earthed, its rod is connected to the body, and the deflection of the needle shows how much the body’s potential differs from the Earth’s (the Earth’s potential is usually taken as zero).

U = φ₁ − φ₂ = A / q, A = q · UU = φ₁ − φ₂ = A / q, A = q · U
where:
  • Upotential difference (voltage) between points 1 and 2, V
  • φ₁, φ₂potentials of the start and end points, V
  • Awork of the field when the charge moves from 1 to 2, J
  • qcharge moved (with its sign), C

1 V is the voltage between two points when the field does 1 J of work moving 1 C of charge. A positive charge released in a field moves from high to low potential, a negative charge (an electron) from low to high potential — in both cases the field does positive work.

Example 5: work and voltage

1) A 3 µC charge is moved from a point at 500 V to a point at 200 V. Find the work of the field.
2) An electron goes from a point at 100 V to a point at 300 V. How much work does the field do?
3) The field does 6 · 10⁻⁶ J of work moving a 2 nC charge. What is the voltage between the points?

Show solution
1) U = φ₁ − φ₂ = 500 − 200 = 300 V; A = q · U = 3 · 10⁻⁶ · 300 = 9 · 10⁻⁴ J (0.9 mJ).
2) A = q · (φ₁ − φ₂) = −1.6 · 10⁻¹⁹ · (100 − 300) = +3.2 · 10⁻¹⁷ J. The work is positive: in a field an electron moves towards higher potential by itself.
3) U = A / q = 6 · 10⁻⁶ / (2 · 10⁻⁹) = 3000 V = 3 kV.

In atomic and particle physics energies are tiny, so instead of joules the electronvolt (eV) is used: the energy an electron gains from the field when it passes through a potential difference of 1 V. For larger energies keV (10³ eV), MeV (10⁶ eV) and GeV (10⁹ eV) are used.

1 eV = 1.6 · 10⁻¹⁹ C · 1 V = 1.6 · 10⁻¹⁹ J
where:
  • eVelectronvolt, a unit of energy
  • eelementary charge, 1.6 · 10⁻¹⁹ C

A particle with charge e starting from rest gains U eV after passing through a voltage U: 100 V → 100 eV. An α-particle with charge 2e gains twice as much at the same voltage — 2U eV.

Example 6 (coded answer): the electronvolt

An α-particle (charge 2e) at rest passes through a potential difference of 500 V. Calculate the kinetic energy it gains, in electronvolts.
Extra questions: a) how many joules is this energy? b) how many joules is 2 keV? c) how many electronvolts is 4.8 · 10⁻¹⁹ J?

Show solution
Ek = q · U = 2e · 500 V = 1000 eV. Answer: 1000.
a) 1000 · 1.6 · 10⁻¹⁹ = 1.6 · 10⁻¹⁶ J.
b) 2000 · 1.6 · 10⁻¹⁹ = 3.2 · 10⁻¹⁶ J.
c) 4.8 · 10⁻¹⁹ / (1.6 · 10⁻¹⁹) = 3 eV.
Typical mistake: taking the α-particle’s charge as e and writing 500.

For a uniform field we now have two expressions for the work: A = q · E · d and A = q · U. Setting them equal gives the link between field strength and voltage. It also explains the second unit of field strength: 1 N/C = 1 V/m. Moving along the field lines, the potential drops by E volts every metre.

E = U / dE = U / d
where:
  • Estrength of the uniform field, V/m
  • Uvoltage between two points (for example, two plates), V
  • ddistance between these points along the field lines, m

For a uniform field only. The field points in the direction in which the potential decreases. Using this formula in a capacitor is covered in the lesson «Capacitance and capacitors».

Example 7: E = U / d

1) Parallel plates are 5 mm apart with 200 V between them. Find the field strength.
2) In a uniform field of 3000 V/m, what is the voltage between two points on the same field line 4 cm apart?
3) In dry air a spark starts at a field strength of about 3 · 10⁶ V/m. What voltage is needed for a spark across a 1 cm air gap?

Show solution
1) d = 5 mm = 5 · 10⁻³ m; E = U / d = 200 / (5 · 10⁻³) = 4 · 10⁴ V/m.
2) U = E · d = 3000 · 0.04 = 120 V.
3) U = E · d = 3 · 10⁶ · 0.01 = 3 · 10⁴ V = 30 kV.

The set of points with the same potential is called an equipotential surface. Moving a charge along such a surface gives U = 0, so the field does no work. That is only possible if the force is perpendicular to the surface: field lines are always perpendicular to equipotential surfaces.

  • The equipotential surfaces of a point charge are concentric spheres centred on the charge (circles in the drawing).
  • The equipotential surfaces of a uniform field are parallel planes perpendicular to the field lines.
  • Field lines point towards decreasing potential; where surfaces drawn at equal steps (say every 10 V) are crowded, the field is strong.
  • The surface (and the whole volume) of a conductor is equipotential.
++q50403020100+−+−+−+−MN12field linesequipotential surfaces
Field lines (arrows) are perpendicular to equipotential surfaces (dashed lines). The numbers on the right are the potentials of the surfaces (V). Moving a charge from M to N along path 1 or path 2 takes the same work of the field — Example 8.
Example 8 (multiple choice): work along different paths

In the uniform field of the drawing a charge q = 2 µC is moved from point M (potential 40 V) to point N (potential 10 V), first along path 1 (a straight line), then along path 2 (a broken line). Which statements are true?
I. The work of the field is 60 µJ on both paths.
II. The work on path 2 is greater because that path is longer.
III. If the charge is brought back from N to M, the total work of the field round the closed path is zero.
A) I only B) II only C) I and III D) II and III E) I, II and III

Show solution
I. U = 40 V − 10 V = 30 V; A = q · U = 2 · 10⁻⁶ · 30 = 6 · 10⁻⁵ J = 60 µJ — true.
II. The electrostatic field is conservative: the work does not depend on the shape or length of the path — false. The vertical parts of path 2 run along equipotential surfaces, where the work is zero; all the work is done on the horizontal part, going from 40 V to 10 V.
III. On the way back the field does −60 µJ, so the total round the closed path is zero — true.
Answer: C.

Conductors and dielectrics in an electrostatic field

A conductor (a metal) contains free electrons. When it is brought into a field, the electrons shift until the resulting field inside the conductor is exactly zero (electrostatic induction; see the lesson «Electric charge, Coulomb’s law and the electric field»). If any field remained inside, the electrons would keep moving. Once equilibrium is reached:

  1. The field strength inside the conductor is zero: E = 0.
  2. All excess charge sits on the outer surface of the conductor; the inner surface of a hollow sphere stays uncharged.
  3. All points of the conductor have the same potential: it is an equipotential body, and field lines meet its surface at right angles.
  4. Charge gathers more densely on sharply curved, pointed parts of the surface — the pointed tip of a lightning rod works this way.

This gives the idea of electrostatic shielding: inside a metal box or a fine metal mesh (a Faraday cage) there is no external electrostatic field. Sensitive instruments are covered by metal screens, and people inside a metal-bodied car or a plane are protected when lightning strikes.

φ = k · q / R (r ≤ R), φ = k · q / r (r ≥ R)φ = k · q / R (r ≤ R), φ = k · q / r (r ≥ R)
where:
  • qcharge of the sphere, C
  • Rradius of the conducting sphere, m
  • rdistance from the centre of the sphere to the point, m

For a charged conducting sphere in air. Outside the sphere the field is as if all the charge sat at the centre. Inside E = 0, and the potential is the same as on the surface. The field strength is greatest just outside the surface: E = k · q / R².

61218900450300φr6121815Er
A charged metal sphere in air (R = 6 cm): φ in V, E in kV/m, r in cm. Inside, the potential is constant and the field is zero; outside φ ~ 1/r and E ~ 1/r².
Example 9 (multiple choice and coded): φ(r) and E(r) graphs of a sphere

The graphs above belong to a charged metal sphere in air.
1) Which answer is correct for the points 3 cm and 18 cm from the centre of the sphere?
A) 3 cm: φ = 900 V, E = 0; 18 cm: φ = 300 V, E ≈ 1.7 kV/m
B) 3 cm: φ = 0, E = 0; 18 cm: φ = 300 V, E ≈ 1.7 kV/m
C) 3 cm: φ = 1800 V, E = 60 kV/m; 18 cm: φ = 300 V, E ≈ 1.7 kV/m
D) 3 cm: φ = 900 V, E = 15 kV/m; 18 cm: φ = 300 V, E = 5 kV/m
E) 3 cm: φ = 900 V, E = 0; 18 cm: φ = 100 V, E ≈ 1.7 kV/m
2) Calculate the charge of the sphere (nC).

Show solution
1) 3 cm < 6 cm — the point is inside the sphere: E = 0, and the potential equals the surface value, φ = 900 V.
18 cm = 3 · 6 cm — outside: φ ~ 1/r → 900 / 3 = 300 V; E ~ 1/r² → 15 / 3² ≈ 1.7 kV/m. Answer: A.
B sets the potential inside to zero as well, C applies k · q / r inside too, D takes E inside equal to its surface value and uses E ~ 1/r, E makes the potential fall as 1/r².
2) At the surface φ = k · q / R → q = φ · R / k = 900 · 0.06 / (9 · 10⁹) = 6 · 10⁻⁹ C. Answer: 6.
Check: E(R) = k · q / R² = 9 · 10⁹ · 6 · 10⁻⁹ / 0.06² = 1.5 · 10⁴ V/m = 15 kV/m — as on the graph.

A dielectric has no free charges, but the field polarises its molecules: polar molecules (water, for example) turn along the field, while in non-polar molecules the electron clouds shift slightly. The bound charges that appear on the dielectric’s surfaces create a field opposite to the external one and weaken the total field — not down to zero as in a conductor, but by a factor of ε.

ε = E₀ / Eε = E₀ / E
where:
  • εrelative permittivity (dielectric constant) of the substance, no unit
  • E₀field strength the same charges create in a vacuum, V/m
  • Efield strength in the dielectric, V/m

ε = 1 for a vacuum, ≈ 1 for air, ≈ 2 for kerosene and paraffin, ≈ 7 for glass, 81 for water. In a dielectric the field strength and potential of a point charge, and the force between charges, are ε times smaller than in a vacuum.

Example 10: potential in a dielectric

1) Find the potential and the field strength 6 cm from a 4 nC point charge in kerosene (ε = 2).
2) In air the potential at a point in the field of a point charge is 810 V. If the whole system is put into water (ε = 81) with the charge and the point unchanged, what will the potential be?
3) An uncharged metal plate and a glass plate of the same size are placed in a uniform field. Inside which one is the field zero?

Show solution
1) φ = k · q / (ε · r) = 9 · 10⁹ · 4 · 10⁻⁹ / (2 · 0.06) = 36 / 0.12 = 300 V; E = k · q / (ε · r²) = 36 / (2 · 0.0036) = 5000 V/m.
2) φ ~ 1/ε → 810 / 81 = 10 V.
3) The metal plate: its free electrons shift and cancel the field inside completely. In glass the field is only weakened: E = E₀ / 7.

Motion of charged particles in an electric field

An electric field can speed up a charged particle or change its direction of motion — X-ray tubes, the cathode-ray tube of an oscilloscope and particle accelerators all work this way. The work of the field equals the change in the particle’s kinetic energy (the work–energy theorem). For electrons and protons gravity is billions of times weaker than the electric force, so it is neglected.

|q| · U = m · v² / 2 − m · v₀² / 2, v = √(2 · |q| · U / m)|q| · U = m · v² / 2 − m · v₀² / 2, v = √(2 · |q| · U / m)
where:
  • |q|magnitude of the particle’s charge, C (e for an electron)
  • Uaccelerating potential difference the particle passes through, V
  • mmass of the particle, kg (electron 9.1 · 10⁻³¹ kg, proton 1.67 · 10⁻²⁷ kg)
  • v₀, vinitial and final speed, m/s; the second formula is for v₀ = 0

The speed is proportional to √U: four times the voltage gives twice the speed and four times the kinetic energy. If the field slows the particle down, the voltage that stops it is U = m · v₀² / (2 · |q|).

Example 11: acceleration by a voltage

1) An electron at rest is accelerated through 182 V. Find its final speed and energy. (e = 1.6 · 10⁻¹⁹ C, m = 9.1 · 10⁻³¹ kg)
2) A proton and an α-particle (charge 2e, mass 4 times the proton’s) start from rest and are accelerated through the same voltage U. What are the ratios of their kinetic energies and speeds?
3) An electron flies into a field at 4 · 10⁶ m/s in the direction of the field lines. Through what potential difference does it pass before it stops?

Show solution
1) v = √(2 · e · U / m) = √(2 · 1.6 · 10⁻¹⁹ · 182 / (9.1 · 10⁻³¹)) = √(6.4 · 10¹³) = 8 · 10⁶ m/s. Energy: Ek = 182 eV = 182 · 1.6 · 10⁻¹⁹ ≈ 2.9 · 10⁻¹⁷ J.
2) Ek = q · U does not depend on mass: the α-particle’s charge is twice as large → its energy is 2 times larger.
v = √(2 · q · U / m): the α-particle’s q / m is 2 / 4 = 1/2 of the proton’s → the speed ratio is √(1/2) ≈ 0.71 (the α-particle is √2 times slower).
3) The force on the electron points against E, so the field slows it down: e · U = m · v₀² / 2 → U = m · v₀² / (2e) = 9.1 · 10⁻³¹ · 16 · 10¹² / (3.2 · 10⁻¹⁹) = 45.5 V.

Now let the particle enter parallel to the plates, i.e. perpendicular to the field lines, with speed v₀. Along the plates no force acts on it, so that motion is uniform: x = v₀ · t. Perpendicular to the plates there is a constant acceleration a = |q| · E / m. It is exactly like a horizontally thrown body, with a in place of g, so the path is a parabola. After leaving the plates the particle flies in a straight line.

a = |q| · U / (m · d), t = l / v₀, y = a · t² / 2 = |q| · U · l² / (2 · m · d · v₀²)a = |q| · U / (m · d), t = l / v₀, y = a · t² / 2 = |q| · U · l² / (2 · m · d · v₀²)
where:
  • aacceleration perpendicular to the plates, m/s²
  • Uvoltage between the plates, V
  • ddistance between the plates, m
  • llength of the plates, m
  • v₀initial speed of the particle parallel to the plates, m/s
  • ydeflection at the exit from the plates (displacement along the field), m

A particle entering midway between the plates misses them if y < d / 2. At the exit the velocity component perpendicular to the plates is a · t, and the deflection angle of the velocity is given by tan β = a · t / v₀.

Interactive
Loading simulation…
Path of a particle between the plates: x is the distance along the plates (cm), y₁ the deflection (cm), y₂ = d / 2 the level of a plate (the particle enters midway). U is the deflecting voltage, U0 the voltage that accelerated the particle beforehand (V), d the gap between the plates (cm). The formula follows from |q| · U0 = m · v₀² / 2, and neither charge nor mass is left in it. For U = 100, U0 = 1000 and d = 2, y = 0.3125 cm at x = 5 cm. If the lines cross, the particle hits a plate.
Example 12 (written task): an electron deflected between plates

An electron moving at 2 · 10⁷ m/s enters exactly midway between plates 5 cm long, parallel to them. The plates are 2 cm apart with 91 V between them (e = 1.6 · 10⁻¹⁹ C, m = 9.1 · 10⁻³¹ kg).
a) Explain why the electron’s path between the plates is a parabola.
b) Calculate the electron’s acceleration, its time between the plates and its deflection at the exit. Will it hit a plate?
c) By what angle is its velocity turned from the original direction when it leaves the plates?

Show solution
a) There is no force along the plates — that motion is uniform (x = v₀ · t). Perpendicular to the plates a constant force F = e · E acts — that motion is uniformly accelerated (y = a · t² / 2). Hence y ~ x², so the path is a parabola. The electron bends towards the positive plate.
b) E = U / d = 91 / 0.02 = 4550 V/m; a = e · E / m = 1.6 · 10⁻¹⁹ · 4550 / (9.1 · 10⁻³¹) = 8 · 10¹⁴ m/s².
t = l / v₀ = 0.05 / (2 · 10⁷) = 2.5 · 10⁻⁹ s.
y = a · t² / 2 = 8 · 10¹⁴ · (2.5 · 10⁻⁹)² / 2 = 2.5 · 10⁻³ m = 2.5 mm.
d / 2 = 10 mm > 2.5 mm — the electron does not hit a plate.
c) Perpendicular velocity: a · t = 8 · 10¹⁴ · 2.5 · 10⁻⁹ = 2 · 10⁶ m/s; tan β = 2 · 10⁶ / (2 · 10⁷) = 0.1 (β ≈ 5.7°).
Check with the shortcut: the electron gets this speed from U₀ = m · v₀² / (2e) = 1137.5 V; y = U · l² / (4 · d · U₀) = 91 · 0.05² / (4 · 0.02 · 1137.5) = 2.5 · 10⁻³ m.

Two parallel plates carrying equal and opposite charges form a device that stores charge and field energy — a capacitor. Its capacitance, connections and energy are in the next lesson, «Capacitance and capacitors».

Key points

  • The electrostatic field is conservative: its work does not depend on the path and is zero round a closed path; in a uniform field A = q · E · d.
  • Potential φ = Wp / q₀ is the scalar energy characteristic of a field; for a point charge φ = k · q / (ε · r), and the potentials of several charges add with their signs.
  • Voltage U = φ₁ − φ₂ = A / q; 1 eV = 1.6 · 10⁻¹⁹ J; in a uniform field E = U / d (1 V/m = 1 N/C).
  • Field lines are perpendicular to equipotential surfaces and point towards decreasing potential; no work is done along an equipotential surface.
  • Inside a conductor E = 0, the charge is on the outer surface and the potential is the same everywhere; a dielectric weakens the field ε times (ε = E₀ / E).
  • Acceleration: |q| · U = m · v² / 2 − m · v₀² / 2; between plates a particle is deflected along a parabola: y = |q| · U · l² / (2 · m · d · v₀²).

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
What is the SI unit of electric potential?