Electric Field – SS1 Physics Lesson Note (Third Term)

An electric field is the region of space around an electric charge in which any other charge placed there experiences an electrical force.

Behavioural objectives

By the end of the lesson, students should be able to:

1. state the fundamental law of electrostatics and Coulomb’s law;
2. define electric field intensity, electric potential and potential difference, and state their units;
3. define the electron volt and convert energy between electron volts and joules;
4. define capacitance and state the factors that affect it;
5. calculate the combined capacitance of capacitors in series and in parallel;
6. calculate the energy stored in a capacitor.

What are the properties of an electric field?

An electric field is a vector quantity, so it has both size and direction.

• Region of force: any electric charge placed in the field feels an electrical force.
• Vector quantity: the field has magnitude and direction at every point.
• Test charge: a small positive charge used to find the direction of the field.
• Direction rule: the field points the way a positive test charge would be pushed.

How are electric field lines drawn?

Electric field lines are lines with arrows that show the direction of the force on a small positive test charge.

• Positive charge: the lines point away from it.
• Negative charge: the lines point towards it.

What is the fundamental law of electrostatics?

The fundamental law of electrostatics states that like charges repel and unlike charges attract.

• Like charges: positive and positive, or negative and negative, push each other apart.
• Unlike charges: a positive and a negative charge pull towards each other.

What is Coulomb’s law?

Coulomb’s law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

• Proportionality: F ∝ q₁q₂ / r²
• Equation: F = k q₁q₂ / r²
• Constant: k = 1 / (4πε₀) = 8.99 × 10⁹ N m² C^-2
• Full form: F = q₁q₂ / (4πε₀r²)
• Symbols: F is the force (N), q₁ and q₂ are the charges (C), r is the distance apart (m).

What is electric field intensity (strength)?

Electric field intensity, E, at a point is the force experienced by a unit positive test charge placed at that point.

• Formula: E = F / q
• Symbols: E is the field intensity (N C^-1), F is the force (N), q is the test charge (C).
• SI unit: newton per coulomb (N C^-1).
• Type: vector quantity.

How is E found for a point charge Q?

1. The force between Q and the test charge q is F = Qq / (4πε₀r²).
2. Write E = F / q = [Qq / (4πε₀r²)] × (1 / q).
3. Cancel q to get E = Q / (4πε₀r²).

What is electric potential?

The electric potential (V) at a point is the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.

• Formula: V = W / q
• Symbols: V is the potential (volts), W is the work done (joules), q is the charge (coulombs).
• Unit: volt (V).
• Type: scalar quantity.
• Point charge: at distance r from a charge Q, V = Q / (4πε₀r).

What are the sign rules for potential?

• Work done against the field: the potential is positive.
• Work done by the field: the potential is negative.
• Zero points: the potential at infinity is zero, and the potential of the earth is also zero.

How is the earth used to test the potential of a body?

A body is connected to the earth by a wire (the body is earthed), and the direction of electron flow shows its potential.

• Electrons flow from the body to the earth: the body is at a negative potential.
• Electrons flow from the earth to the body: the body is at a positive potential.
• Comparison: positive points are at higher potential, and negative points are at lower potential.

What is potential difference?

The potential difference between two points in an electric field is the work done in taking a unit positive charge from one point to the other.

• Two potentials: if a charge Q moves from a point at V₁ to a point at V₂, the p.d. is (V₁ – V₂).
• Work done: W = Q(V₁ – V₂)

What is the relation between electric field intensity and potential difference?

For a uniform field, the p.d. between two points equals the field intensity multiplied by the distance between them.

1. Let a charge Q move from A to B through a distance X.
2. Work done = force × distance, so W = F X.
3. Since E = F / Q, the force is F = EQ.
4. Substitute to get W = EQX.
5. Also W = Q(V_a – V_b), so Q(V_a – V_b) = EQX.
6. Cancel Q to get V_a – V_b = EX.
7. Therefore E = (V_a – V_b) / X, that is E = p.d. / distance, or V = EX.

What is the electron volt (eV)?

The electron volt is the energy gained by an electron in accelerating through a potential difference of one volt.

• Electronic charge: e = 1.6 × 10^-19 C.
• Value in joules: 1 eV = 1.6 × 10^-19 × 1 = 1.6 × 10^-19 J.
• Energy gained: it depends only on the charge and the p.d. through which the particle falls.
• Moving electron: its kinetic energy is ½mv².

What is the kinetic energy of an electron moving in a circle?

1. The centripetal force towards the centre is mv²/r.
2. The electrical force of attraction, e² / (4πε₀r²), provides it.
3. Equate them: mv²/r = e² / (4πε₀r²).
4. Multiply by r: mv² = e² / (4πε₀r).
5. Halve both sides: ½mv² = ½ × e² / (4πε₀r).

Worked example: energy of an α particle

Calculate the energy in eV and joules of an α particle (helium nucleus) accelerated through a p.d. of 4 × 10⁶ V.

1. The charge on an α particle is 2e.
2. KE = work done = charge × p.d. = qV.
3. KE = 2 × 4 × 10⁶ = 8 × 10⁶ eV = 8 MeV.
4. Use 1 eV = 1.6 × 10^-19 J.
5. KE = 8 × 10⁶ × 1.6 × 10^-19 = 1.28 × 10^-12 J.

What is a capacitor?

A capacitor is a device for storing electrical energy or charge.

• General form: any two conductors that are electrically insulated from their surroundings.
• Common type: two parallel plate conductors separated by a very small distance, d.
• Charging: connect the plates across a battery, and they carry equal and opposite charges.
• Potential difference: the p.d. across the plates is V.
• Circuit symbol: —-| |—-

What is capacitance?

The capacitance of a capacitor is the ratio of the charge Q on either conductor to the potential difference V between the two conductors.

• Formula: C = Q / V, so Q = CV.
• SI unit: the farad (F), equal to one coulomb per volt (C V^-1).
• Parallel plate capacitor: C = εA / d.
• Symbols: A is the plate area, d is the plate separation, ε is the permittivity of the dielectric (F m^-1).

What factors affect capacitance?

Three factors affect the capacitance of a parallel plate capacitor.

1. Area of the plates: a larger area gives a larger capacitance.
2. Separation of the plates: a smaller separation gives a larger capacitance.
3. Dielectric between the plates: a dielectric of higher permittivity gives a larger capacitance.

Worked example: finding capacitance

A capacitor holds a charge of 4.0 × 10^-4 C when a p.d. of 400 V is applied. Calculate its capacitance.

1. C = Q / V.
2. C = (4.0 × 10^-4) / 400.
3. C = 1.0 × 10^-6 F = 1.0 μF.

How do capacitors combine in series and in parallel?

The net capacitance of a group of capacitors depends on whether they are joined in series or in parallel.

• Series: capacitors joined end to end in one path, so the same charge is on each.
• Parallel: capacitors joined side by side across the same p.d.

• Compare with resistors: the rules for resistors are the opposite of these.

Worked example: two equal capacitors

Find the net capacitance of two 4 μF capacitors joined (a) in series and (b) in parallel.

1. Series: 1/C = 1/4 + 1/4 = 1/2, so C = 2 μF.
2. Parallel: C = 4 + 4 = 8 μF.

How much energy is stored in a capacitor?

A charged capacitor stores electrical potential energy equal to ½QV, which is also ½Q²/C or ½CV².

1. Moving a charge Q through a p.d. does work W = average p.d. × charge = ½QV.
2. Put V = Q / C to get W = ½ × (Q / C) × Q = ½Q²/C.
3. Put Q = CV to get W = ½CV².
4. This work is stored in the capacitor as electrical potential energy.

Evaluation

1. State Coulomb’s law.
2. Define electric field intensity and state its SI unit.
3. Define the electron volt and state its value in joules.
4. A capacitor stores a charge of 6.0 × 10^-4 C when a p.d. of 200 V is applied across it. Calculate its capacitance.
5. List three factors that affect the capacitance of a parallel plate capacitor.

Assignment

1. Two capacitors of 3 μF and 6 μF are connected (a) in series and (b) in parallel. Find the net capacitance in each case. Then calculate the energy stored in the parallel combination when it is charged to a p.d. of 100 V.