Cambridge AS & A Level9702

Electric fields

Physics 9702 Chapter Notes

What this chapter covers

Electric fields - Electric fields and field linesElectric fields - Uniform electric fieldsElectric fields - Electric force between point chargesElectric fields - Electric field of a point chargeElectric fields - Electric potential
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In this chapter

  1. 1Introducing Electric Fields
  2. 2Coulomb's Law
  3. 3Radial Electric Fields
  4. 4Electric Potential
  5. 5Uniform Electric Fields
  6. 6Electric vs. Gravitational Fields

1. Introducing Electric Fields

An electric field is a region of space around an electric charge where another charge would experience a force. It's a way to describe how a charge influences the space around it. We visualise these fields using 'electric field lines'. The rules are simple: 1) Field lines show the direction of the force on a positive test charge. This means they point away from positive charges and towards negative charges. 2) The closer the field lines are to each other, the stronger the electric field. 3) Field lines never cross. For a single point charge, the field lines radiate outwards (positive) or inwards (negative). For two parallel plates with a potential difference, the field between them is uniform, meaning the field lines are parallel, equally spaced, and point from the positive plate to the negative plate.

E = F / q

Key term

Electric Field: A region of space where an electric charge experiences a non-gravitational force.

Examiner insight

Examiners award marks for clear diagrams showing correct line direction (with arrows), appropriate spacing to show field strength, and lines that start and end on charges or plates correctly.

Common pitfall

Drawing field lines that cross each other. Since the field at any point has a single unique direction, field lines can never cross.

Worked example 14 marks

Draw the electric field pattern fora) a single positive point charge, andb) two parallel metal plates with the top plate at +100V and the bottom plate at 0V.

  1. 1

    a) For a single positive point charge, draw a circle representing the charge with a '+' inside. Draw at least 8 straight lines radiating outwards from the surface of the circle. Add arrows to each line pointing away from the charge.

  2. 2

    b) Draw two horizontal parallel lines. Label the top one '+100V' and the bottom one '0V'. In the space between the plates, draw at least 4 vertical, parallel, equally spaced lines. Add an arrow to each line pointing downwards, from the positive plate towards the zero-volt plate.

Recap

  • An electric field is a region where a charge feels a force.
  • Electric field lines show the path a positive test charge would take.
  • Field lines point away from positive charges and towards negative charges.
  • The density of field lines represents the strength of the field.
  • Field lines between parallel plates are parallel and equally spaced, indicating a uniform field.

Quick check

  1. What is the direction of the electric field at a point?1 mark
  2. What does the spacing of electric field lines indicate?1 mark

2. Coulomb's Law

Coulomb's Law describes the force between two stationary point charges. It states that the force is directly proportional to the product of the two charges (Q1 and Q2) and inversely proportional to the square of the distance(r) between them. The formula is F = (Q1 * Q2) / (4πε₀r²). Here, ε₀ (epsilon-nought) is a fundamental constant called the permittivity of free space, approximately 8.85 x 10⁻¹² F m⁻¹. A positive result for F indicates a repulsive force (like charges), while a negative result indicates an attractive force (opposite charges). This 'inverse square law' is a crucial concept in physics, also seen in gravitation.

F = (Q₁Q₂) / (4πε₀r²)

Key term

Coulomb's Law: The law stating that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Common pitfall

Forgetting to square the distance 'r' in the denominator, or using distance in cm instead of converting to meters.

Fun fact

The electric force between a proton and an electron in a hydrogen atom is about 10³⁹ times stronger than the gravitational force between them!

Worked example 14 marks

Two point charges, Q₁ = +2.0 nC and Q₂ = -5.0 nC, are separated by a distance of 4.0 cm. Calculate the magnitude and nature of the force between them. (ε₀ = 8.85 x 10⁻¹² F m⁻¹)

  1. 1

    Step 1: Convert all units to SI. Q₁ = +2.0 x 10⁻⁹ C, Q₂ = -5.0 x 10⁻⁹ C, r = 4.0 cm = 0.040 m.

  2. 2

    Step 2: Write down Coulomb's Law. F = (Q₁Q₂) / (4πε₀r²).

  3. 3

    Step 3: Substitute the values into the formula. F = ((2.0 x 10⁻⁹) * (-5.0 x 10⁻⁹)) / (4 * π * 8.85 x 10⁻¹² * (0.040)²).

  4. 4

    Step 4: Calculate the numerator. (2.0 x 10⁻⁹) * (-5.0 x 10⁻⁹) = -1.0 x 10⁻¹⁷ C².

  5. 5

    Step 5: Calculate the denominator. 4 * π * 8.85 x 10⁻¹² * (0.040)² = 1.778 x 10⁻¹⁶ F m.

  6. 6

    Step 6: Calculate the final force. F = (-1.0 x 10⁻¹⁷) / (1.778 x 10⁻¹⁶) = -5.6 x 10⁻² N.

  7. 7

    Step 7: State the magnitude and nature. The magnitude of the force is 5.6 x 10⁻² N. The negative sign indicates that the force is attractive.

Recap

  • Coulomb's Law calculates the force between two point charges.
  • The force is proportional to the product of the charges (Q₁Q₂).
  • The force is inversely proportional to the square of the distance (1/r²).
  • A positive force is repulsive; a negative force is attractive.
  • Always convert distances to metres and charges to Coulombs before calculating.

Quick check

  1. If the distance between two charges is tripled, what happens to the magnitude of the force between them?2 marks

3. Radial Electric Fields

While Coulomb's Law gives the force between two specific charges, we often want to describe the field created by a single charge, Q. The electric field strength (E) at a distance(r) from a point charge Q is defined as the force per unit positive charge (E = F/q). By substituting Coulomb's Law for F, we get E = [ (Qq)/(4πε₀r²) ] / q. The test charge 'q' cancels out, leaving the formula for the electric field strength of a point charge: E = Q / (4πε₀r²). This field is called a radial field because it radiates outwards from (or inwards towards) the central charge. Like the force, the field strength follows an inverse square law with distance.

E = Q / (4πε₀r²)

Key term

Electric Field Strength (E): The force per unit positive charge experienced by a small test charge placed at a point in the field.

Examiner insight

Candidates must be able to derive the formula for E from Coulomb's Law and the definition of field strength (E=F/q). Marks are often awarded for showing this link clearly.

Common pitfall

Confusing the formula for electric field strength (E ∝ 1/r²) with the formula for electric potential (V ∝ 1/r).

Worked example 13 marks

A small sphere carries a charge of +6.8 μC. Calculate the electric field strength at a distance of 15 cm from its centre.

  1. 1

    Step 1: Convert units to SI. Q = +6.8 μC = +6.8 x 10⁻⁶ C. r = 15 cm = 0.15 m.

  2. 2

    Step 2: Write down the formula for electric field strength of a point charge. E = Q / (4πε₀r²).

  3. 3

    Step 3: Substitute the values. E = (6.8 x 10⁻⁶) / (4 * π * 8.85 x 10⁻¹² * (0.15)²).

  4. 4

    Step 4: Calculate the denominator. 4 * π * 8.85 x 10⁻¹² * (0.15)² ≈ 2.50 x 10⁻¹².

  5. 5

    Step 5: Calculate the final field strength. E = (6.8 x 10⁻⁶) / (2.50 x 10⁻¹²) = 2.72 x 10⁶ N C⁻¹.

  6. 6

    Step 6: State the direction. Since the source charge is positive, the field is directed radially outwards.

Recap

  • The electric field from a point charge is a radial field.
  • Field strength E is calculated using E = Q / (4πε₀r²).
  • Field strength is a vector; its direction is the direction of force on a positive charge.
  • E is proportional to the source charge Q and inversely proportional to the square of the distance r.
  • This formula applies outside a charged conducting sphere as if all charge were at its centre.

Quick check

  1. What are the standard units for electric field strength?1 mark
  2. How does the electric field strength at 20 cm from a charge compare to the strength at 10 cm?2 marks

4. Electric Potential

Electric potential (V) at a point in an electric field is the work done per unit charge in bringing a positive test charge from infinity to that point. It's a measure of the electric potential energy per coulomb. Unlike electric field strength, potential is a scalar quantity – it has magnitude but no direction. The formula for the potential at a distance r from a point charge Q is V = Q / (4πε₀r). Note that potential is proportional to 1/r, not 1/r² like field strength. Potential is positive around a positive charge and negative around a negative charge. The potential at infinity is defined as zero. The work done (W) in moving a charge q between two points is the charge multiplied by the change in potential (ΔV), so W = qΔV.

V = W / q

V = Q / (4πε₀r)

W = qΔV

Key term

Electric Potential (V): The work done per unit positive charge in bringing a small test charge from infinity to a point in an electric field.

Examiner insight

Questions often require calculating the change in potential energy when a charge moves between two points. Students must be able to calculate the potential at each point and then find the difference.

Common pitfall

Mixing up electric potential (a scalar, in Volts, ∝ 1/r) with electric field strength (a vector, in N C⁻¹, ∝ 1/r²). They are different concepts with different formulas.

Worked example 15 marks

A point charge of -4.5 nC is in a vacuum.a) Calculate the electric potential at a point P, 0.25 m away from the charge.b) Calculate the work done in moving a proton (charge +1.6 x 10⁻¹⁹ C) from infinity to point P.

  1. 1

    a) Step 1: Write down the formula for electric potential. V = Q / (4πε₀r).

  2. 2

    a) Step 2: Substitute the values. Q = -4.5 x 10⁻⁹ C, r = 0.25 m. V = (-4.5 x 10⁻⁹) / (4 * π * 8.85 x 10⁻¹² * 0.25).

  3. 3

    a) Step 3: Calculate the potential. V = -161.8 V. So, V ≈ -162 V.

  4. 4

    b) Step 1: Use the definition of potential V = W/q, so W = qV.

  5. 5

    b) Step 2: The potential at infinity is 0V. The potential at P is -162 V. The work done is moving the proton from infinity to P.

  6. 6

    b) Step 3: Substitute values. q = +1.6 x 10⁻¹⁹ C, V = -162 V. W = (1.6 x 10⁻¹⁹) * (-162).

  7. 7

    b) Step 4: Calculate the work done. W = -2.59 x 10⁻¹⁷ J. The negative sign means the field does work, so energy is released.

Recap

  • Electric potential is the electric potential energy per unit charge.
  • Potential is a scalar quantity, measured in Volts (V) or Joules per Coulomb (J C⁻¹).
  • For a point charge, potential V is given by V = Q / (4πε₀r).
  • Potential is positive for positive charges and negative for negative charges.
  • The work done to move a charge q through a potential difference ΔV is W = qΔV.

Quick check

  1. What is the value of the electric potential at an infinite distance from any charge?1 mark
  2. Is electric potential a scalar or a vector quantity?1 mark

5. Uniform Electric Fields

A uniform electric field is one where the field strength is the same in magnitude and direction at all points. The best way to create one is with two large, parallel conducting plates separated by a small distance, with a potential difference (voltage) V applied across them. The field lines are parallel, equally spaced, and point from the positive plate to the negative one. The electric field strength, E, is related to the potential difference V and the plate separation d by the simple formula E = V/d. Because the field is uniform, the force on a charge q placed in the field is constant (F = qE). This constant force causes the charge to have a constant acceleration (a = F/m = qE/m). If a charge enters this field at right angles to the field lines, it will follow a parabolic path, similar to a projectile in a uniform gravitational field.

E = V / d

F = qE

Key term

Uniform Electric Field: An electric field in which the field strength is constant in both magnitude and direction at all points, typically found between two parallel charged plates.

Examiner insight

Examiners frequently test the motion of charged particles in uniform fields, often combining it with mechanics (F=ma) and kinematics to calculate velocity, time, or deflection.

Common pitfall

Applying the formula E = V/d to radial fields around a point charge. This formula is only valid for uniform fields.

Worked example 15 marks

Two parallel plates are separated by 3.0 cm and have a potential difference of 1.5 kV between them. An electron (m = 9.11 x 10⁻³¹ kg, q = -1.6 x 10⁻¹⁹ C) is released from rest at the negative plate. Calculate its acceleration towards the positive plate.

  1. 1

    Step 1: Convert units to SI. d = 3.0 cm = 0.030 m. V = 1.5 kV = 1500 V.

  2. 2

    Step 2: Calculate the electric field strength. E = V / d = 1500 / 0.030 = 50,000 V m⁻¹ (or N C⁻¹).

  3. 3

    Step 3: Calculate the electric force on the electron. The magnitude of the force is F = |q|E. F = (1.6 x 10⁻¹⁹ C) * (50,000 N C⁻¹) = 8.0 x 10⁻¹⁵ N.

  4. 4

    Step 4: Use Newton's second law (F=ma) to find the acceleration. a = F / m.

  5. 5

    Step 5: Substitute values. a = (8.0 x 10⁻¹⁵ N) / (9.11 x 10⁻³¹ kg) = 8.78 x 10¹⁵ m s⁻².

  6. 6

    Step 6: State the final answer. The acceleration of the electron is 8.8 x 10¹⁵ m s⁻² towards the positive plate.

Recap

  • A uniform electric field exists between two parallel charged plates.
  • Field lines are parallel, equally spaced, and point from high potential to low potential.
  • Field strength is constant and given by E = V/d.
  • The force on a charge q in a uniform field is constant: F = qE.
  • A charged particle experiences constant acceleration in a uniform field.

Quick check

  1. Two plates are 5 cm apart with a 200 V potential difference. What is the electric field strength?2 marks

6. Electric vs. Gravitational Fields

Electric fields and gravitational fields share some remarkable similarities but also have crucial differences. Both are examples of fields of force that act over a distance, and both follow an inverse square law for the force between two point sources (masses or charges). However, the source of a gravitational field is mass, while the source of an electric field is charge. A key difference is that mass is always positive, so gravity is always attractive. Charge, however, can be positive or negative, meaning the electric force can be either attractive or repulsive. This allows for more complex interactions and phenomena like shielding, which is not possible with gravity.

Force Law: F = Gm₁m₂/r² vs F = Q₁Q₂/(4πε₀r²)

Field Strength: g = GM/r² vs E = Q/(4πε₀r²)

Potential: φ = -GM/r vs V = Q/(4πε₀r)

Key term

Inverse Square Law: A law stating that a specified physical quantity is inversely proportional to the square of the distance from the source of that quantity.

Examiner insight

Marks are awarded for identifying both similarities (e.g., inverse square law, radial field shape) and fundamental differences (e.g., source of field, nature of force).

Fun fact

If you could remove all the electrons from two 1-gram paperclips and hold them one meter apart, the repulsive force between the now positively charged paperclips would be strong enough to lift a weight equal to the entire Earth's population.

Worked example 14 marks

State two similarities and two differences between the electric field of a point charge and the gravitational field of a point mass.

  1. 1

    Similarity 1: Both fields are radial, with field lines pointing outwards from or inwards towards the centre.

  2. 2

    Similarity 2: The strength of both fields obeys an inverse square law with distance from the point source (E ∝ 1/r² and g ∝ 1/r²).

  3. 3

    Difference 1: The electric force can be either attractive or repulsive, as charge can be positive or negative. The gravitational force is always attractive, as mass is always positive.

  4. 4

    Difference 2: The electric field acts on charges, whereas the gravitational field acts on masses. It is possible to shield an object from an electric field, but not from a gravitational field.

Recap

  • Both fields obey an inverse square law for force and field strength.
  • Both fields have a potential that varies as 1/r.
  • Gravitational fields are created by mass; electric fields are created by charge.
  • Gravitational forces are always attractive; electric forces can be attractive or repulsive.
  • The gravitational force is significantly weaker than the electric force.
  • Electric fields can be shielded, but gravitational fields cannot.

Quick check

  1. What property of matter is the source of a gravitational field?1 mark
  2. Can a gravitational field be repulsive?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define electric field strength and state its SI unit.2 marks
  2. Two point charges of +4.0 nC and -6.0 nC are placed 8.0 cm apart in a vacuum. Calculate the magnitude of the force between them.3 marks
  3. A proton produces a radial electric field. Calculate the electric field strength at a distance of 0.50 x 10⁻¹⁰ m from the centre of the proton. (Charge of a proton = +1.60 x 10⁻¹⁹ C)3 marks
  4. Two large parallel plates are separated by 2.5 cm and have a potential difference of 5.0 kV. a) Calculate the electric field strength between the plates. b) Calculate the force on an oil drop carrying a charge of +3.2 x 10⁻¹⁹ C that is in the field.4 marks
  5. a) Calculate the electric potential at a distance of 4.0 cm from the centre of a sphere carrying a charge of +5.0 nC. b) Calculate the work done in bringing an electron (charge -1.6 x 10⁻¹⁹ C) from a very distant point to this point.4 marks
  6. Two point charges, A (+20 nC) and B (-5 nC), are placed on a line 12 cm apart. Find the position on the line joining A and B where the net electric field is zero.5 marks
  7. An electron is fired with a horizontal velocity of 2.0 x 10⁷ m s⁻¹ into a uniform electric field. The field has a strength of 1.0 x 10⁴ N C⁻¹ and is directed vertically downwards. The field exists over a horizontal length of 5.0 cm. Calculate the vertical deflection of the electron as it exits the field.6 marks
  8. State two ways in which electric potential differs from electric field strength.2 marks
  9. A conducting sphere of diameter 8.0 cm is charged to a potential of +500 V. a) Calculate the charge on the sphere. b) Calculate the electric field strength at the surface of the sphere.5 marks
  10. Compare and contrast electric fields and gravitational fields. You should refer to the sources of the fields, the nature of the forces, and the mathematical forms of the field strengths.6 marks

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