Electricity & Magnetism

Electrostatics

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to state and apply Coulomb's Law to calculate the force between point charges.
  • By the end of this lesson students will be able to define electric field strength and calculate it for a point charge.
  • By the end of this lesson students will be able to define electric potential (HL) and calculate it for a point charge.
  • By the end of this lesson students will be able to distinguish between electric field strength (vector) and electric potential (scalar).

Key concepts

Coulomb's Law

Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centres. The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

F = (1 / 4πε₀) * (q₁q₂ / r²) or F = k * (q₁q₂ / r²)
Electric Field Strength (E)

The electric field strength at a point is defined as the force per unit positive charge experienced by a small test charge placed at that point. It is a vector quantity, with its direction being the direction of the force on a positive test charge. Electric field lines are used to represent electric fields; they originate from positive charges and terminate on negative charges, and their density indicates the strength of the field.

E = F / q or E = (1 / 4πε₀) * (Q / r²) or E = k * (Q / r²)
Electric Potential (V) (Higher Level)

The electric potential at a point in an electric field is defined as the work done per unit positive charge in bringing a small test charge from infinity to that point. It is a scalar quantity. The unit of electric potential is the Volt (V), which is equivalent to Joules per Coulomb (J C⁻¹).

V = W / q or V = (1 / 4πε₀) * (Q / r) or V = k * (Q / r)

Key facts to remember

  • 1Coulomb's Law describes the electrostatic force between point charges.
  • 2The constant k (1 / 4πε₀) has a value of 9 x 10⁹ N m² C⁻² in a vacuum.
  • 3Electric field strength (E) is a vector quantity, representing force per unit positive charge (N C⁻¹ or V m⁻¹).
  • 4Electric potential (V) is a scalar quantity, representing work done per unit positive charge (V or J C⁻¹).
  • 5Like charges repel, and unlike charges attract.
  • 6Electric field lines point away from positive charges and towards negative charges.
  • 7Always convert units to SI units (metres for distance, Coulombs for charge) before calculation.

Worked examples

Example 1

Two point charges, +3.0 μC and -5.0 μC, are placed 15 cm apart in a vacuum. Calculate the magnitude of the electrostatic force between them.

IIdentify given values: q₁ = +3.0 μC = +3.0 x 10⁻⁶ C, q₂ = -5.0 μC = -5.0 x 10⁻⁶ C, r = 15 cm = 0.15 m.
IIState the constant: k = 9 x 10⁹ N m² C⁻².
IIIWrite down Coulomb's Law formula for magnitude: F = k |q₁q₂| / r².
IVSubstitute values: F = (9 x 10⁹) * |(3.0 x 10⁻⁶) * (-5.0 x 10⁻⁶)| / (0.15)².
VCalculate: F = (9 x 10⁹) * (15 x 10⁻¹²) / 0.0225.
VIF = (135 x 10⁻³) / 0.0225 = 6.0 N.

Answer

6.0 N

The force is attractive because the charges are of opposite signs.

Example 2

Calculate the electric field strength at a point 10 cm from a point charge of +2.0 μC in a vacuum.

IIdentify given values: Q = +2.0 μC = +2.0 x 10⁻⁶ C, r = 10 cm = 0.10 m.
IIState the constant: k = 9 x 10⁹ N m² C⁻².
IIIWrite down the formula for electric field strength due to a point charge: E = k Q / r².
IVSubstitute values: E = (9 x 10⁹) * (2.0 x 10⁻⁶) / (0.10)².
VCalculate: E = (18 x 10³) / 0.01 = 1.8 x 10⁶ N C⁻¹.

Answer

1.8 x 10⁶ N C⁻¹

The electric field is directed away from the positive charge.

Example 3

Calculate the electric potential at a point 20 cm from a point charge of +4.0 μC in a vacuum. (Higher Level)

IIdentify given values: Q = +4.0 μC = +4.0 x 10⁻⁶ C, r = 20 cm = 0.20 m.
IIState the constant: k = 9 x 10⁹ N m² C⁻².
IIIWrite down the formula for electric potential due to a point charge: V = k Q / r.
IVSubstitute values: V = (9 x 10⁹) * (4.0 x 10⁻⁶) / 0.20.
VCalculate: V = (36 x 10³) / 0.20 = 1.8 x 10⁵ V.

Answer

1.8 x 10⁵ V

Electric potential is a scalar quantity. If the charge Q were negative, the potential V would also be negative.

Common mistakes

  • Forgetting to convert units (e.g., cm to m, μC to C) before using formulas.
  • Using r instead of r² in Coulomb's Law or the electric field strength formula.
  • Confusing electric field strength (vector) with electric potential (scalar).
  • Incorrectly handling the sign of charges, especially in electric potential calculations where the sign of the charge determines the sign of the potential.
  • Not showing all steps in calculations, which can lead to loss of marks in exams.

Exam tips

  • Always list the known quantities and the unknown quantity you need to find. This helps in selecting the correct formula.
  • Show all steps in your calculations clearly. Even if your final answer is incorrect, you may get marks for correct method.
  • Pay close attention to units and ensure your final answer has the correct SI units.
  • For problems involving multiple charges, drawing a clear diagram can help visualise the forces or fields and their directions.

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