Electrostatics

This section explores electric field concepts and Coulomb’s law of electrostatics to understand how electric forces between charges depend on their magnitude and separation. We will study electric field strength, which measures the intensity at a point in an electric field, and potential difference, which reflects the energy needed to move a charge between two points. Capacitors, which store energy by holding charge on their plates as well as their arrangement and behaviour will be examined in both DC and AC circuits focusing on charging and discharging. Understanding how capacitors store and manage energy is essential for applications like power supplies, signal filtering, and energy storage, as well as for understanding how electronic devices and electric fields function in real-world scenarios like camera flashes, defibrillators (used to provide high voltages to start the heart), touchscreens and sensors, etc.

KEY IDEAS

• A capacitor is a charge storing device.

• Capacitance is the ratio of the charge to the potential to which it is raised.

• Charging a capacitor is the process of depositing charges on the plates of a capacitor when it is connected to a DC source.

• Coulomb’s law of electrostatics states that: “the magnitude of 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 them”.

• Discharging a capacitor is a process in which the voltage across the capacitor decreases due to the flow of charges through a resistor, which has a voltage across it proportional to the current flowing through it.

• Electric field is the region or area around a charged body where the electric force of the charged body can be experienced.

• Electric field strength is the force per unit positive charge at a point.

• Potential difference is the work done in moving a unit charge from one point to another in an electric field.

Electric Field Patterns

When a magnetic material is placed near a permanent magnet, it experiences a magnetic force because it is within the magnet’s magnetic field. Similarly, an electric field can either attract or repel charges depending on their nature. Just as a magnet’s field draws magnetic materials toward it, the charges influenced by the electric field experience an electric force because they are located within that field.

An electric field is the area or region around a charge where the electric force of the charge can be experienced. An electric field is represented with electric field lines of electric lines of force.

Some characteristics of electric field lines

1. The direction of the electric field at any point is indicated by the direction of the field lines at that point.

2. Electric field lines never intersect or cross each other.

3. Closer lines indicate a stronger electric field.

4. Electric field lines point away from positive charges and toward negative charges.

5. Electric field lines meet surfaces at 90 degrees to the surface.

6. For two like charges, the field lines repel and curve outward.

7. For two opposite charges, the lines attract and connect the two charges directly.

A: Electric field lines of a positive charge B: Electric field lines of a negative charge C: Electric field lines of positive charges repelling D: Electric field lines of negative charges repelling E: Electric field lines of opposite charges attracting

Figure 3.1: Electric field lines

Activity 3.1 Exploring Electric Fields

Materials Needed

1. Balloons (preferably two of different colours)

2. Small pieces of lightweight paper (cut or ripped into small squares)

3. String (to hang the balloons)

4. A clear area to conduct the experiment What to do

1. Inflate the two balloons and tie them securely.

2. Cut the lightweight paper into small squares to serve as “charged particles.”

3. Hang the balloons from a string in a space where they can move freely, ensuring they are about 1-2 feet apart.

4. Rub each balloon on your hair or a piece of fabric to give them static charge.

5. Slowly bring the two charged balloons close together and record your observation.

6. Now, take the small pieces of papers and gently sprinkle them around the balloons.

Figure 3.2: Papers sprinkled on balloon

7. Record your observation.

8. Identify friends in your class and discuss your observations from the

activity in relation to these questions:

a. How did the balloons behave when brought close together?

b. What happened to the confetti (small pieces of coloured paper) near the charged balloons?

Activity 3.2 Exploring Electric Field Patterns through Video Research Objective: To understand electric field patterns by researching and watching videos that demonstrate the visualisation of these patterns using semolina on an oil surface.

Materials Needed

1. Access to the internet (computers, tablets, or smartphones)

2. Video platforms (YouTube, educational websites)

3. Notebook and pen/pencil (for taking notes) What to do

1. In small groups, review what you know about electric fields.

2. Use your device (computer, tablet, or smartphone) to search for videos that demonstrate the visualisation of electric field patterns using semolina on an oil surface. Here are some suggested search terms:

a. “Electric field visualisation semolina oil”

b. “Electric field patterns demonstration”

c. “Physics experiments with electric fields”

3. Choose at least two videos to watch. Look for videos that clearly explain the setup and the results of the experiment.

4. Watch the videos but as you watch each video, pay close attention to and take notes on:

a. the materials used in the experiment.

b. how the semolina behaves in response to the electric field.

c. any explanations provided about the principles behind what you are observing.

d. key observations about the electric field patterns.

e. important scientific concepts or terminologies mentioned.

f. any questions or thoughts that arise during the viewing.

5. After watching, prepare to share your findings with your classmates.

Think about:

a. what you found most interesting or surprising.

b. how this experiment helps you understand electric fields better.

c. any differences you noticed between the videos you watched.

6. Participate in a group discussion where you will share your observations and insights with your classmates. Be ready to discuss your notes and answer questions from others.

7. After the discussion, write a short reflection (1-2 paragraphs) about what you learned from this activity. Consider how visual demonstrations can enhance your understanding of abstract concepts like electric fields.

Activity 3.3 Experiment to Visualise Electric Fields Using Semolina Materials Needed

• A large flat tray or shallow dish

• Castor oil

• Semolina (or fine sand)

• Two metal objects (e.g., two metal plates, spheres, nails etc)

• A high-voltage power supply

• A piece of cardboard or stiff paper

• Electrical wires with alligator clips

• A ruler (optional) Procedure

1. Place the flat tray or shallow dish on a stable surface. Fill it with a thin layer of castor oil, which has been warmed on a radiator, spreading it evenly across the bottom.

2. Position the two metal objects in the tray, one near each side. These will serve as the electrodes. Ensure that both ends are submerged in the castor oil.

3. Attach electrical wires with alligator clips to the metal objects. Connect the wires to the high-voltage power supply.

4. Sprinkle a thin later of iron filings over the top.

5. Set the high-voltage power supply to a low voltage first. Gradually increase the voltage. As you increase the voltage, the electric field created by the metal objects will begin to influence the semolina on the tray. The semolina grains will shift and settle into patterns that reflect the shape of the electric field. These patterns are formed because the grains are small and lightweight, and they move in response to the electric forces.

6. Sketch a diagram of what you see.

7. Try changing the distance between the electrodes to see how it affects the field pattern.

8. Test different shapes and sizes of the electrodes (e.g., using spheres or flat plates) to see how they influence the field lines.

Tips

• Make sure the voltage is not too high to avoid any risk of electrical shock.

• The patterns may be clearer if you gently tap the tray to settle the semolina into position.

Activity 3.4 Hypothesising Electric Field Patterns

For each of the diagrams below, sketch your prediction of the shape and direction of the electric field pattern that would be produced between the two charged objec

Figure 3.3: Arrangements of Charged Objects

Coulomb’s law of electrostatics

Figure 3.4: Charles Coulomb

Coulomb’s Law is all about how electric charges interact with each other. As learnt earlier, when you have two balloons and you rub them on your hair, they get static electricity and can stick to each other or even repel (push away) from each other. Coulomb’s Law helps us understand why that happens!

The Basics of Charges

There are two types of electric charges: positive (+) and negative (−)

1. Like Charges Repel: If you have two positive charges or two negative charges, they push away from each other.

2. Opposite Charges Attract: If you have one positive charge and one negative charge, they pull towards each other.

Coulomb’s Law tells us how strong the force is between two charges. Here’s how it works:

1. Distance Matters: The closer the charges are to each other, the stronger the force. Imagine if you and your friend are trying to hug.

2. Amount of Charge Matters: The more charge something has, the stronger the force.

Coulomb’s law of electrostatics states that: “the magnitude of 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 them” Mathematically, this can be expressed as:

F ∝ |^(q)1 × q₂|______ r²F = k |^(q)1 × q₂|_ r²Where:

• F is the electrostatic force

• k is Coulomb’s constant

• q₁and q₂are the magnitudes of the charges

• r is the distance between the charges Coulomb’s constant, k, is equal to 1/4πε0 .

ε₀is known as the permittivity of free space, a constant which reflects the ability of an electric field to pass through a vacuum.

Activity 3.5 Finding the unit for Coulomb’s constant and for permittivity of free space By analysing the units in the formula for Coulomb’s law (above), find the unit for Coulomb’s constant. Note: you should start by rearranging the formula to make k the subject. Choose the correct answer from the options below:

1. m²C⁻²2. NmC⁻²3. Nm²C⁻²4. Nm²C² Next, use this result to find the unit for permittivity of free space.

Activity 3.6 Exploring Coulomb’s Torsion Balance and Electrostatic

Forces Objective: To learn about Coulomb’s torsion balance and discover the relationship between electrostatic force, distance, and charge.

Figure 3.5: Coulomb’s Torsion Balance

What You Will Do

1. Find and watch the video titled “Coulomb’s Torsion Balance” on YouTube.

2. While watching the video, take notes on the following points:

a. How does the torsion balance measure electrostatic force?

b. What happens to the angle of twist when charges are introduced?

c. How do distance and charge affect the electrostatic force?

3. After watching the video, you will participate in a group discussion with your classmates. Share your notes and observations. Discuss these questions:

a. What did you find most interesting about Coulomb’s experiments?

b. Why is understanding electrostatic forces important in physics?

c. Can you think of real-world applications of Coulomb’s findings?

4. Write a short reflection in your notebook about what you learned from this activity. Include:

a. Your understanding of the relationship between charge, distance, and electrostatic force.

b. How Coulomb’s work has influenced modern physics.

Activity 3.7 Calculating Electrostatic Force

Read the worked examples below before attempting the example questions that follow.

Worked Example 1

Two point charges of +3 μC and -5 μC are separated by a distance of 2 metres in a vacuum. Calculate the magnitude of the force between them. (Use Coulomb’s constant k = 9 × 10⁹Nm²/C².)

Step-by-Step Solution for Question 1

Step 1: Identify the given values Charge 1, q₁= +3 μC = 3 × 10⁻⁶C Charge 2, q₂= -5 μC = -5 × 10⁻⁶C Distance between the charges, r = 2 m Coulomb’s constant, k = 9 × 10⁹N·m²/C²

Step 2: Introduce the Coulomb’s Law

F = k |^(q)1 × q₂|_ r²Step 3: Substitute the known values into the formula F = 9 × 10⁹3 × 10⁻⁶× 5 × 10⁻⁶_______________ 2²Step 4: Simplify and calculate to obtain the answer F = 9 × 10⁹1.5 × 10⁻¹¹_________ 4 F = 3.375 × 10⁻²N

Worked Example 2

Three charges are positioned along a straight line. +1 μC is at the origin, -3 μC is 2 metres to the right, and +4 μC is 3 metres to the right of the +1 μC charge.

Calculate the net force on the +1 μC charge. (Use k = 9 × 10⁹Nm²/C²).

Step-by-Step Solution

Step 1: Identify the given values:

Charge 1 (q₁): +1 μC = +1 × 10⁻⁶C (at origin) Charge 2 (q₂): -3 μC = -3 × 10⁻⁶C (2 m to the right) Charge 3 (q₃): +4 μC = +4 × 10⁻⁶C (3 m to the right) Distance between q₁and q₂(r₁₂): 2 m Distance between q₁and q₃(r₁₃): 3 m Coulomb’s constant (k): 9 × 10⁹N·m²/C²

Step 2: Sketch the diagram to show the distribution of charges on a straight line

Step 3: Calculate the force between q₁ and q₂ (F₁₂) F₁₂= k |^(q)1 × q₂|______ r₁₂² F₁₂= 9 × 10⁹| + 1 × 10 −6 × − 3 × 10 −6 |_________________ 2²F₁₂ = 9 × 10⁹3 × 10⁻¹²_ 4 F₁₂= 6.75 × 10-³ N Since q₁ and q₂ have opposite charges, the force F₁₂ is attractive, and it acts towards the right.

Step 4: Calculate the force between q₁ and q₃ (F₁₃) F₁₃ = k |q 1 × q 3|_ r₁₃² F₁₃ = 9 × 10⁹| + 1 × 10 −6 × + 4 × 10 −6 |_________________ 3²F₁₃ = 9 × 10⁹4 × 10⁻¹²_ 9 F₁₃= 4 × 10-³ N Since q₁ and q₃ have the same charge, the force F₁₃ is repulsive, and it acts towards the left.

Step 5: Determine the net force on q₁ Since F₁₂ acts towards the right and F₁₃ acts towards the left, we subtract the smaller force from the larger one to find the net force:

Net force = F₁₂- F₁₃ Net force = 6.75 × 10-³ N - 4 × 10-³ N Net force = 2.75 × 10-³ N Therefore, the net force on the +1 μC charge is 2.75 × 10-³ N, directed towards the right.

Practice Problems

Now, using the worked example as a guide, solve the following problems individually or in groups.

1. Two point charges, +2 μC and -4 μC, are separated by a distance of 3 metres in a vacuum. Calculate the magnitude of the electrostatic force between them. (Use k = 9 × 10⁹N·m²/C²).

2. Two positive charges of +5 μC each are placed 5 metres apart in a vacuum.

Determine the magnitude of the repulsive force between them. (Use k = 9 × 10⁹N·m²/C²).

3. Two point charges, +4 μC and -6 μC, exert an attractive force of 0.3 N on each other in a vacuum. Calculate the distance between the charges. (Use k = 9 × 10⁹N·m²/C²).

4. Three charges are positioned along a straight line. A +2 μC charge is at the origin, a -5 μC charge is 3 metres to the right, and a +6 μC charge is 4 metres to the right of the +2 μC charge. Calculate the net force on the +2 μC charge.

(Use k = 9 × 10⁹N·m²/C²).

Activity 3.8 Design an experiment to verify Coulomb’s Law Design an experiment which could be used to verify Coulomb’s inverse-square law; i.e. to prove that the force acting on a charged particle in an electric field decreases with distance according to 1/r².

You are not expected to conduct your experiment.

Activity 3.9 Research task Conduct research on the use of Coulomb’s Law in technology. Prepare a report and presentation explaining its importance in devices such as electrostatic precipitators and photocopiers.

Electric field strength We know that a body raised above the ground level has a certain amount of gravitational potential energy which, by definition, is given by the amount of work done in raising it to that height. The body falls because there is attraction due to gravity. Its fall always proceeds from a place of higher potential energy to one of lower potential energy.

Now, consider an electric field. Imagine an isolated positive charge Q placed in air.

Like Earth’s gravitational field, it has its own electrostatic field which theoretically extends up to infinity. If the charge q is very far away from Q, say, at infinity, then the force on it is zero. As q is brought nearer to Q, a force of repulsion acts on it (as similar charges repel each other), hence work or energy is required to bring it to a point within in the electric field (point A).

Hence, when at point A, charge q has some amount of electric potential energy.

Similar other points in the field will also have some potential energy.

Figure 3.7: Source charge Q with field E and test charge q The strength of the field at any point is defined as the force experienced by a unit positive charge placed at that point. Its direction is the direction along which the force acts.

Mathematically, the field strength can be determined as; E = F __ q measured in NC⁻¹, Where:

E is the electric field strength F is the force of attraction or repulsion in newton (N) q is charge in coulomb (C) but F = k | Q × q|_ r² Then the electric field strength (E) due to a point charge (Q) at a distance r is given by:

E = k|Q|_ r²Electric potential (V) It is the work done (W) or energy spent on bringing a unit test charge from infinity to a point in an electric field, i.e., the work done per unit charge in moving from infinity to a point in an electric field.

Mathematically stated, V = W/q , in J C⁻¹or volts You may wonder, if a charge is already in the field, and is moving between any two points within the field, does it have to do any work? Certainly yes, just the way you do some work in moving from your classroom to the school field. The work done by a unit charge is the difference in electric potential between the two points. It is called the potential difference, or voltage.

For example, if a unit charge moves from point A to another point B, the work done, or potential difference ∆ V = V_(A) − V_(B) Relationship between E and V: Uniform electric fields Consider two parallel conducting plates, A and B, separated by d,

Figure 3.8: Parallel conducting plates The potential at A is V_(A)and that at B is V_(B).

E = F __ q = ( W _ d )___ q = W__ qd = V __ d E = V __ d ⌈E = V/d = V_(AB)___ d ⌉, measured in Vm⁻¹ For a charge q moving from A to B, the potential difference is:

V_(AB) = V_(B) − V_(A) = ∆ V V_(AB) = −(V_(A) − V_(B)) = ∆ V V_(AB) = (V_(A) − V_(B)) = − ∆ V Where V_(AB) = ∆ V in magnitude Therefore, the work done W, in moving the charge is:

w = Fd, but F = Eq, and Ed = V W = Eqd W = Vq, but V = − ∆ V = V_(AB) W = − q ∆ V W = qV_(AB)

Activity 3.10 Differences between Electric Potential and Electric Field Intensity Fill in the table below

Table 3.1: Electric Potential and Electric Field Intensity

Electric Potential Electric Field Intensity

Definition Unit

Scalar or vector Variation with distance from source charge

Activity 3.11 How Electric Field Strength and Potential Vary with Distance in a Uniform Field Materials needed

1. Mathematical set

2. Calculator

3. At least four graph sheets What to do

1. Form groups of three, copy and complete the table below using the following information:

E = V/d , given that E = 8.99 × 10⁷Vm⁻¹.

d/m 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 V/V

2. Plot two graphs:

a. E on the vertical axis and d on the horizontal axis

b. V on the vertical axis and d on the horizontal axis

3. Compare your graphs with the solutions in Annex A.

4. Explain how the graphs confirm the relationship established in the equations above.

Activity 3.12 How electric field strength and potential vary with distance Materials needed

1. Mathematical set

2. Calculator

3. At least four graph sheets What to do

1. Form groups of three, copy and complete the tables below using the following information:

E = k Q_ r²…… … … 1, k = 8.99 × 10⁹Nm²C⁻², Q = 1.0 μC V = k Q/r ………..2 r/m 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 r²/m² E/ NC⁻¹ V/V

2. Plot two graphs

a. E on the vertical axis and r on the horizontal axis

b. V on the vertical axis and r on the horizontal axis

3. Compare your graphs with the solutions in Annex A.

4. Explain how the graphs confirm the relationship established in the equations [1 and 2] above.

Activity 3.13 Calculating Electric field strength and Electric potential Read the worked examples below before attempting the example questions that follow.

Worked Example 1

Two points, A and B, are in an electric field created by a point charge. The electric potential at point A is V_(A)=120 V, and at point B, it is V_(B)= 80 V. What is the potential difference between points A and B?

Step-by-Step Solution for Question 1

Step 1: Identify the given values V_(A)=120 V V_(B)= 80 V

Step 2: Introduce the potential difference ΔV formula ΔV=V_(A)−V_(B)

Step 3: Substitute the given values:

ΔV = 120 V−80 V

Step 4: Calculate to get your answer ΔV = 120 V−80 V =40 V The potential difference between points A and B is 40 V

Worked Example 2

A uniform electric field exists between two parallel plates separated by a distance of 0.05 m. If the potential difference between the plates is 200 V, calculate the electric field strength between the plates.

Step-by-Step Solution

Step 1: Identify the given values d = 0.05 m V = 200 V

Step 2: Introduce the formula for electric field strength E E = V/d

Step 3: Substitute the given values:

E = 200/0.05

Step 4: Calculate to get your answer E = 4000 V / m The electric field strength between the plates is 4000 V/m.

Worked Example 3

A 2×10⁻⁶C charge is moved between two points in an electric field where the potential difference is 50 V. How much work is done to move this charge?

Step-by-Step Solution

Step 1: Identify the given values q = 2×10⁻⁶C V = 200 V

Step 2: Introduce the formula for work done W to move a charge q across a potential difference V W= qV

Step 3: Substitute the given values W = 2×10⁻⁶×50

Step 4: Calculate to get your answer W = 1×10⁻⁴J The work done to move the charge is 0.0001 J or 1×10⁻⁴J

Worked Example 4

A charge +5 μC is placed in a uniform electric field 2×10⁴N/C. Calculate the force experienced by the charge.

Step-by-Step Solution

Step 1: Identify the given values q = 5μC = 5×10⁻⁶C E = 2×10⁴N/C.

Step 2: Recall the formula for electric field strength and make force the subject E = F/q F = qE

Step 3: Substitute the given values F = (5 × 10⁻⁶) × (2 × 10⁴)

Step 4: Calculate to get your answer F = 0.1 N

Worked Example 5

Two charges, +3 μC and -2 μC , are separated by a distance of 0.5 m. Calculate the electric field strength at a point midway between them. The Coulomb constant is k = 9 × 10⁹Nm²/C².

Step-by-Step Solution

Step 1: Recall the formula for the electric field due to a point charge E = k|Q|____ r² Where:

• E = electric field strength (in N/C),

• k = Coulomb constant (9 × 10⁹N·m²/C²),

• Q = magnitude of the charge (in C),

• r = distance from the charge to the point of interest (in m).

Step 2: Identify the midpoint and distances The point of interest is midway between the charges. Therefore:

r = 0.5/2 = 0.25 m

Step 3: Calculate the electric field due to each charge

a. Electric field due to Q1:

E₁ = k|^(Q)1|_ r²b. Substituting the values:

E₁ = 9 × 10⁹× 2 × 10⁶______________ 0.25²E₁= 4.32 × 10⁵N/C

c. Electric field due to Q2:

E₁ = k|^(Q)2|_ r²d. Substituting the values:

E₁ = 9 × 10⁹× 3 × 10⁶_____________ 0.25²E₂= 2.88 × 10⁵N/C

Step 4: Determine the directions of the fields The electric field due to Q₁(positive charge) points away from Q₁. The electric field due to Q₂(negative charge) points towards Q₂. At the midpoint, both fields point in the same direction (towards Q2) because the charges are oppositely signed.

Step 5: Add the electric fields to et the net electric field strength Since the fields point in the same direction, the net electric field is:

Eₙₑₜ= E₁+ E₂ Eₙₑₜ= (4.32 × 10⁵) + (2.88 × 10⁵) Eₙₑₜ= 7.2 × 10⁵N/C Practice Problems Now, using the worked example as a guide, solve the following problems individually or in groups.

1. Two points, X and Y, are in an electric field created by a point charge. The electric potential at point X is 150 V, and at point Y, it is 90V. What is the potential difference between points X and Y?

2. The potential difference across two parallel plates is 300 V, and they are separated by a distance of 0.03 m. Calculate the electric field strength between the plates.

3. A charge of 5×10⁻⁶C is moved between two points in an electric field where the potential difference is 30 V. Calculate the work done in moving this charge.

4. Two charges, Q₁= +4 μC and Q₂= −3 μC, are separated by a distance of 0.6 m. Calculate the electric field strength at a point midway between them.

The Coulomb constant is k = 9 × 10⁹Nm²/C².

Having a water tank at home is essential for storing water for use when the tap stops running. The tank holds a specific volume of water, which represents its capacity. When water is pumped into the tank, it fills up to a certain level. The pressure of the water being pumped influences how quickly the tank reaches its full capacity. When the tap stops running and you open the tap at the bottom of the tank, water begins to flow out, and eventually, the tank will become empty.

Figure 3.11: Capacitor

A capacitor is essential for storing electrical energy in a circuit, acting as a backup energy source when needed. It has a specific capacitance, which indicates the maximum amount of electric charge it can hold, measured in Farads.

When voltage is applied to a capacitor, it allows electric current to flow in, charging the capacitor until it reaches its maximum capacity. The voltage across the capacitor influences the charging speed; higher voltages can result in a quicker charging process, similar to how higher water pressure fills a water tank more rapidly.

When connected to a circuit that requires energy, the capacitor discharges its stored electric charge, supplying power to the circuit components. Once the capacitor has provided its energy, it becomes discharged and can be recharged, much like refilling a water tank.

Capacitance is defined as the ratio of the charge to the potential to which it is raised.

C = Q/V where C = capacitance, Q = charge, V = voltage or potential difference.

Figure 3.12: Circuit symbol for a capacitor Parallel plate capacitors A parallel plate capacitor is a simple type of capacitor that consists of two parallel conducting plates separated by a dielectric material.

Key Components

• Plates: These are typically made of metal and act as electrodes. They are equal in size and shape.

• Dielectric: This is an insulating material placed between the plates. It can be air, paper, ceramic, or other materials. The dielectric increases the capacitor’s capacitance.

How it works

1. Charging: When a voltage is applied across the plates, one plate becomes positively charged and the other negatively charged.

2. Electric Field: The electric field forms between the plates, storing energy in the electric field.

3. Capacitance: The ability of the capacitor to store charge is measured by its capacitance (C), which depends on the area of the plates (A), the distance between them (d), and the permittivity of the dielectric material (ε):

C = εA/d But ε = ε_(O) εᵣ Where ε_(O) is permittivity of vacuum, εᵣ is relative permittivity C = ε O εᵣ A_ d But for no dielectric (εᵣ = 1) Therefore C = ε O A_ d

Activity 3.14 Modelling Parallel Plate Capacitors

Under the supervision of your teacher, stand up and arrange yourselves into a large circle.

Designate the following role to some individuals:

Person A: Positive terminal of the battery Person B: Negative terminal of the batter Person C: Switch Person D: Positive plate of capacitor Person E: Negative plate of the capacitor (Note: persons A and B should be stood next to one another, as should persons D and E) All other class mates should act as individual electrons able to ‘flow’ when the switch is closed.

Consider the questions below and use these, with your teacher’s guidance, to model the action of the electrons in a circuit containing a capacitor:

1. What is the charge on an electron?

2. Which way will it travel around the circuit, given that the positive plate of the battery is over here (Person A)

3. The charges build up on one side of the capacitor as they reach it. Which side must that be?

4. What kind of material must be between the plates of a capacitor to ensure that the electrons do not ‘jump the gap’ and continue to flow around the circuit?

5. What is happening to the electrons on the other plate of the capacitor?

Why? What charge does this give that plate?

When you have finished your model, draw a diagram to summarise the knowledge you have acquired.

Activity 3.15 Exploring Capacitance with Interactive Simulations

Objective: To use interactive simulations to explore how changing variables like plate area, distance between plates, and dielectric material affects the capacitance of a capacitor.

Materials Needed

1. Computer or Tablet

2. Internet Connection

3. Notebook and Pen/Pencil

What You Will Do

1. Open the PhET Capacitor Lab simulation on your computer or tablet.

Make sure you are familiar with the interface and tools available in the simulation.

Figure 3.13: PhET Capacitor Lab

2. Begin by observing how a basic capacitor works. Adjust the voltage and watch how charges build up on the plates.

3. Experiment with different variables:

a. Plate Area: Change the size of the capacitor plates and observe how this affects capacitance.

b. Distance Between Plates: Increase and decrease the distance between the plates to see its impact on capacitance.

4. As you conduct your experiments, record your observations in your notebook. Note how each variable affects capacitance. Consider using a

table to organise your data clearly.

5. After completing your experiments, analyse your recorded data. Look for patterns or trends that emerge from changing each variable. Answer these questions in your analysis:

a. How does increasing plate area affect capacitance?

b. What happens when you increase the distance between plates?

6. Based on your analysis, draw conclusions about the relationships between plate area, distance, dielectric material, and capacitance.

Activity 3.16 Building a Simple Capacitor with Aluminium Foil and Wax Paper Objective: To learn how to construct a simple capacitor using aluminium foil and wax paper.

Materials Needed

1. Aluminium foil

2. Wax paper (acts as the dielectric)

3. Scissors

4. Ruler

5. Tape

6. Multimeter (for testing capacitance) What to do

1. Find and watch the instructional video titled “How to Build a Simple Capacitor Using Aluminium Foil and Wax Paper.” Pay attention to the steps and safety precautions mentioned.

2. Prepare the Plates: Cut two rectangular pieces of aluminium foil to the same size. Aim for dimensions of around 10 cm by 20 cm; this size will make handling easier.

3. Prepare the Dielectric: Cut a piece of wax paper slightly larger than the aluminium foil pieces (about 12 cm by 22 cm). This wax paper will act as the insulator between the two pieces of foil.

4. Assemble the Capacitor

a. Place one piece of aluminium foil flat on a table.

b. Lay the wax paper directly on top of this first piece of aluminium foil, ensuring it fully covers the foil and extends slightly beyond the edges.

c. Place the second piece of aluminium foil on top of the wax paper, aligning it with the first piece of foil but leaving an edge of wax paper around it to prevent the plates from touching.

5. Secure the Layers: Use tape to hold the edges of the assembly together, making sure the foil pieces don’t touch each other. This ensures the wax paper insulates the foil pieces effectively.

6. Attach Leads (Optional): Use a multimeter to measure the capacitance, attach a lead to each piece of foil to act as contacts.

7. Test the Capacitor

a. If a multimeter with a capacitance setting is available, measure the capacitance by connecting each lead to a piece of foil.

b. Alternatively, connect the capacitor to a low-voltage (1.5V) battery briefly, then discharge it by touching the leads together. You may observe a small spark or feel a slight shock, demonstrating stored charge.

Capacitors in Series and Parallel

Capacitors can be connected in series or parallel to achieve different equivalent capacitances. Let’s explore each configuration:

Capacitors in series Multiple capacitors are said to be connected in series if the negative plate of one capacitor is connected to the positive plate of another capacitor and so on. In this grouping, current is same through each capacitor, the voltage across each capacitor is different and the charge on each capacitor is the same.

Consider 3 capacitors of capacitances C₁, C₂and C₃in series. V is the total voltage supplied. V₁, V₂and V₃are the voltage drops across C₁, C₂and C₃respectively (as shown in the figure below).

Figure 3.14: Capacitors in series V1 = Q/C₁ , V₂ = Q/C₂ , V₃ = Q/C₃ The total pd, V, across the network V = V₁ + V₂ + V₃ V = Q/C₁ + Q/C₂ + Q/C₃ V= Q ( 1/C₁ + 1/C₂ + 1/C₃ ) But V = Q/C Q/C= Q ( 1/C₁ + 1/C₂ + 1/C₃) 1/C = 1/C₁ + 1/C₂ + 1/C₃ Thus, for n capacitors in series, their effective or equivalent capacitance C is given by 1/C = 1/C₁ + 1/C₂ + 1/C₃ + … 1/Cₙ Therefore, the equivalent capacitance is smaller than the smallest individual capacitance. Series connections are used to increase the voltage rating of a capacitor bank.

Capacitors in parallel Multiple capacitors are said to be connected in parallel if the positive plate of each capacitor is connected to the positive terminal of battery and the negative plate of each capacitor is connected to the negative terminal of the battery. In this grouping, the voltage across each capacitor is the same.

Consider 3 capacitors of capacitances C₁, C₂and C₃connected in parallel. V is the supplied voltage. Q₁, Q₂and Q₃are the charges on capacitors C₁, C₂and C₃(as shown in the figure below).

Figure 3.15: Capacitors in Parallel

The sum of the separate charges is given by Q = Q₁ + Q₂ + Q₃ But Q =CV Q₁= C₁V,Q₂= C₂V, Q₃ = C₃ V CV = C₁ V + C₂ V + C₃ V CV = V(C₁ + C₂ + C₃) C= C₁ + C₂ + C₃ + .....Cₙ Thus, for n capacitors in parallel, their effective or equivalent capacitance C is given by C = C₁ + C₂ + C₃ + …Cₙ Therefore, the equivalent capacitance is greater than the largest individual capacitance.

Parallel connections are used to increase the total capacitance of a circuit.

Activity 3.17 Exploring the Arrangement of Capacitors

Objective: Investigate the behaviour of capacitors when arranged in series and parallel configurations on a breadboard, and understand how these arrangements affect total capacitance.

Materials Needed

1. Breadboard

2. 3 Capacitors of different values (e.g., 10 μF, 22 μF, 47 μF)

3. Jumper wires

4. Multimeter (to measure capacitance and voltage)

5. Power supply (e.g., 9V battery)

6. Resistor (e.g., 1 kΩ)

7. Notebook for recording observations

8. Access to a video guide What to do

1. Review how capacitors work and the difference between series and parallel arrangements. Remember:

a. In series, the total capacitance C is given by:

1/C = 1/C₁ + 1/C₂ + 1/C₃

b. In parallel, the total capacitance C is simply the sum:

C = C₁ + C₂ + C₃

2. Search and watch an instructional video that demonstrates how to set up circuits with capacitors in series and parallel. Let the video also guide you through the process step-by-step.

3. Connect the three capacitors in series on the breadboard. Use jumper wires to connect:

a. The positive terminal of the first capacitor to the negative terminal of the second capacitor.

b. The positive terminal of the second capacitor to the negative terminal of the third capacitor.

c. Connect the free positive terminal of the first capacitor to the positive terminal of your power supply.

d. Connect the free negative terminal of the third capacitor to ground (negative terminal) of your power supply.

4. Measure Total Capacitance in Series:

a. Use a multimeter to measure the total capacitance across the series connection. Record this value in your notebook.

b. Compare this measured value with your calculated value using the series formula.

5. Now, connect the same three capacitors in parallel on a new section of the breadboard. Use jumper wires to connect:

a. All positive terminals together and connect them to the positive terminal of your power supply.

b. All negative terminals together and connect them to ground (negative terminal) of your power supply.

6. Measure Total Capacitance in Parallel:

a. Again, use a multimeter to measure the total capacitance across this parallel connection. Record this value in your notebook.

b. Compare this measured value with your calculated value using the parallel formula.

7. Analyse Your Results:

a. Compare your measured capacitances from both configurations with theoretical calculations.

b. Discuss how changing from series to parallel affects total capacitance and what that means for circuit design.

Activity 3.18 Calculating Electrostatic Force

Read the worked examples below before attempting the example questions that follow

Worked Example 1

A capacitor has a charge of 24 μC and a voltage of 12 V across its plates.

Calculate the capacitance of the capacitor.

Step-by-Step Solution for Question 1

Step 1: Identify the given values Charge Q = 24 μC = 24 × 10⁻⁶F Voltage V = 12 V

Step 2: Identify the formula C = Q/V

Step 3: Substitute the given values into the formula C = 24 × 10⁻⁶F/12 V

Step 4: Calculate the capacitance:

C = 2 × 10⁻⁶C = 2 μC

Worked Example 2

Two parallel plates are separated by a distance of 3 mm and have an area of 0.04 m². The space between the plates is filled with air. Calculate the capacitance of the parallel-plate capacitor. (Given ε₀ = 8.85 × 10-¹² F/m) Step-by-Step Solution

Step 1: Identify the given values Given:

A = 0.04 m² d = 3 mm = 3 × 10-³ m ε₀ = 8.85 × 10-¹² F/m

Step 2: Identify the formula for the capacitance of a parallel-plate capacitor C = ε_(O) A_ d

Step 3: Substitute the known values into the formula C = 8.85 ×10⁻¹²F/m ×0.04 m²__________________ 3 ×10⁻³m

Step 4: Calculate to get the answer C = 1.18 × 10-¹⁰F

Worked Example 3

Three capacitors, C₁= 4 μF, C₂= 6 μF, and C₃= 8 μF, are connected in parallel. Calculate the equivalent capacitance of the combination.

Step-by-Step Solution

Step 1: Identify the given values C₁= 4 μF C₂= 6 μF C₃= 8 μF

Step 2: Identify the formula C = C₁ + C₂ + C₃

Step 3: Substitute the given values into the formula C = 4 μF + 6 μF + 8 μF

Step 4: Add the values:

C = 18 μF

Worked Example 4

Three capacitors, C₁= 2 μF, C₂= 5 μF, and C₃= 10 μF, are connected in series. Calculate the equivalent capacitance of the combination.

Step-by-Step Solution

Step 1: Identify the given values C₁= 2 μF C₂= 5 μF C₃= 10 μF

Step 2: Identify the formula 1/C = 1/C₁ + 1/C₂ + 1/C₃

Step 3: Substitute the given values into the formula 1/C = 1/2 μF + 1/5 μF + 1/10 μF

Step 4: Calculate

1/C = 8/10 μ F

Step 5: Take the reciprocal of the result to find the equivalent capacitance C = 10 μF/8 = 1.25 μF Practice Problems Now, using the worked example as a guide, solve the following problems in groups

1. A capacitor is charged to a voltage of 20 V, and it holds a charge of 60 μC.

Calculate the capacitance of the capacitor.

2. A parallel plate capacitor has a plate area of 0.03 m²and a separation distance between the plates of 2×10⁻³m. The dielectric between the plates is air, with a permittivity of free space 8.85×10⁻¹²F/m. Calculate the capacitance of the capacitor.

3. Three capacitors, C₁= 5 μF, C₂= 10 μF, and C₃= 15 μF, are connected in parallel. Calculate the equivalent capacitance of the combination.

4. Three capacitors, C₁= 2 μF, C₂= 4 μF, and C₃= 8 μF, are connected in series. Calculate the equivalent capacitance of the combination.

Activity 3.19 Researching and Presenting Types of Capacitors

Objective: To research different types of capacitors, learn about their structures, functions, and real-world applications.

Materials Needed

1. Access to computers or tablets with internet for research

2. Notebooks and pens for note-taking

3. Presentation software (e.g., PowerPoint, Google Slides)

4. Projector or screen for presentations What to do

1. Get together with 2-3 classmates to form a small group.

2. Decide as a group which type of capacitor you want to research. Here are some options:

a. Ceramic Capacitors

b. Electrolytic Capacitors

c. Film Capacitors

d. Supercapacitors

e. Mica Capacitors

f. Tantalum Capacitors

3. Use the internet to gather information about your chosen capacitor type.

Focus on these key areas:

a. Structure: Describe how the capacitor is built (materials used, design).

b. Function: Explain how it operates and its electrical properties.

c. Applications: Identify where this capacitor is commonly used in real-world devices.

4. Prepare a presentation summarising your findings. Include:

a. Key points about the structure, function, and applications of your capacitor type.

b. Visual aids such as diagrams or images to illustrate your points.

c. Any interesting facts or historical context related to your capacitor type.

5. Take turns presenting your findings to the class. Aim for about 5-7 minutes per group. Be ready to answer questions from your classmates after your presentation.

6. After all presentations, engage in a discussion with your classmates about what you learned. Share thoughts on how different types of capacitors are used in various applications.

Capacitor in a DC Circuit

1. Initial Charging: When a DC voltage is first applied to a capacitor, it starts charging. Current flows through the circuit as the capacitor plates accumulate charge.

2. Steady State: As the capacitor charges, the voltage across its plates increases, opposing the applied voltage. Eventually, the voltage across the capacitor equals the source voltage, and the current flow ceases.

3. Open Circuit Behaviour: Once fully charged, a capacitor acts as an open circuit, blocking the flow of direct current.

Activity 3.20 Building and Analysing a Simple DC Circuit with a Capacitor Objective: To construct a simple DC circuit using a capacitor and resistor, observe the charging and discharging process, measure the voltage across the capacitor over time, and calculate the energy stored in the capacitor.

Materials Needed

1. Breadboard

2. 1 DC power supply (battery or power adapter)

3. 1 Resistor (e.g., 10 kΩ)

4. 1 Capacitor (e.g., 4200 μF)

5. Multimeter (to measure voltage)

6. Stopwatch or timer

7. Notebook for recording data

8. Graph paper What to do

1. Set Up the Circuit: Connect the circuit as follows:

a. Place the resistor and capacitor on the breadboard.

b. Connect one terminal of the resistor to the positive terminal of the DC power supply.

c. Connect the other terminal of the resistor to one terminal of the capacitor.

d. Connect the other terminal of the capacitor to the negative terminal of the DC power supply, completing the circuit.

e. Ensure that you identify and connect the capacitor correctly, noting its polarity if it is a polarised capacitor.

2. Charging Phase

a. Close the switch or connect the battery to start charging the capacitor.

b. Use the multimeter to measure and record the voltage across the capacitor at regular intervals (e.g., every 10 seconds) for about 3 minutes or until it reaches a steady value.

c. Record your voltage measurements in your notebook.

3. Discharging Phase

a. After charging, disconnect the power supply.

b. Now, connect a short wire across the terminals of the capacitor to discharge it.

c. Again, measure and record the voltage across the capacitor at regular intervals as it discharges. Continue measuring until the voltage drops close to zero.

4. Using your recorded voltage values during both charging and discharging, create a table with time (in seconds) and corresponding voltage readings.

5. On graph paper, plot your voltage readings against time for both charging and discharging phases. You should see exponential curves representing both processes.

6. Analyse your graphs. Discuss how well they match theoretical expectations regarding exponential growth during charging and decay during discharging.

Capacitor in an AC Circuit

1. Alternating Current: In an AC circuit, the voltage across the capacitor continuously changes polarity.

2. Charging and Discharging: As the AC voltage increases in one direction, the capacitor charges. As the voltage decreases and reverses polarity, the capacitor discharges. This continuous charging and discharging process results in an alternating current flow through the capacitor.

3. Capacitive Reactance: Capacitors oppose the flow of alternating current.

This opposition is known as capacitive reactance (Xc), which is inversely proportional to the frequency of the AC signal and the capacitance of the capacitor.

Energy stored in a capacitor A capacitor stores energy in the form of an electric field between its plates.

When a voltage is applied to the capacitor, it causes a separation of charges, with positive charges accumulating on one plate and negative charges on the other. This separation of charges creates an electric field between the plates.

The energy stored in the capacitor is equal to the work done in separating these charges. This energy can be expressed in several equivalent ways:

• E = 1/2 CV²• E = 1/2 QV

• E = Q²___ 2C Where:

• E is the energy stored in the capacitor (in Joules)

• C is the capacitance of the capacitor (in Farads)

• V is the voltage across the capacitor (in Volts)

• Q is the charge stored on the capacitor (in Coulombs) This stored energy can be released later by discharging the capacitor, which allows the charges to flow back together. This release of energy can be used for various applications, such as powering flashlights, providing energy for electronic circuits, or even delivering a high-energy shock in a defibrillator.

Activity 3.21 Revisiting Activity 3.20 – Calculating Energy stored in a Capacitor

1. Look back at the data you collected during Activity 3.20.

a. Use the formula for energy stored in a capacitor:

E = 1/2 C V²where:

E is energy in Joules, C is capacitance in Farads, V is voltage across the capacitor at each time point.

b. Calculate E for various measured voltages during both charging and discharging phases.

Activity 3.22 Exploring Capacitor Applications

Objective: To research and discover how capacitors are used in various electronic applications.

Materials Needed

1. Access to the internet for research

2. Notebook or digital document for recording your findings

3. Blank table template (see below) What to do

1. Review what a capacitor is and its basic function in electronic circuits.

2. Use the internet to find examples of where and how capacitors are used.

Focus on at least three different applications. Here are some areas to explore:

a. Power supplies

b. Filters (such as audio or RF filters)

c. Timing circuits

d. Energy storage systems

e. Tuning circuits

3. As you research, fill in the table below with details about each application you discover.

Table 3.3: Capacitor Usage

Application Name Description of use Type of Capacitor Used Benefits

4. Once you have completed your research and filled in the table, be ready to share your findings with your classmates. Highlight interesting facts and discuss why capacitors are important in each application.

E = 1/2 CV²Activity 3.23 Calculating Energy stored in Capacitor Read the worked example below before attempting the example questions that follow.

Worked Example

A capacitor with a capacitance of 4 μF is charged to a voltage of 50 V. Calculate the energy stored in the capacitor.

Step-by-Step Solution

Step 1: Identify the given values Capacitance, C=4 μF =4 ×10⁻⁶F Voltage, V=50 V

Step 2: Identify the formula E = 1/2 CV²Step 3: Substitute the given values into the formula E = 1/2 4 ×10⁻⁶×50²Step 4: Calculate E = 1/2 4 ×10⁻⁶×2500 E = 0.005 J Practice Problems Now, using the worked example as a guide, solve the following problems in groups

1. A capacitor of 5 μF is charged to a voltage of 10 V. Calculate the energy stored in the capacitor.

2. A capacitor stores 2 mJ of energy when connected to a 20 V source. Calculate the capacitance of the capacitor.

3. A capacitor holds a charge of 30 μC at a voltage of 15 V. Calculate the energy stored in the capacitor.

Activity 3.24 Revisit Activity 3.17 – Exploring the arrangement of capacitors Set up your experiment again as in Activity 3.17.

Use the formulae for energy stored on a capacitor to compare the total energy stored in both series and parallel arrangements compared to the energy stored by the cumulative energy stored by the constituent capacitors connected individually to the same power supply.

Activity 3.25 Disassembling Old Electronics

Many electronic devices contain capacitors. If you have any such devices at home, have a go at disassembling them to identify the location of the capacitor.

Can you suggest the purpose of the inclusion of a capacitor in the design?

• Asiedu, P., & Baah-Yeboah, H. A. (n.d.). Physics for Senior High Schools (4th ed., pp. 385-399). Aki-Ola Publications.

• Halliday, D., Resnick, R., & Walker, J. (2018). Fundamentals of Physics (11th ed., pp. 599-613, 609-622, 630-650). John Wiley & Sons.

• Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers with Modern Physics (9th ed., pp. 568-580, 741-754). Brooks/Cole Cengage Learning.

• Giancoli, D. C. (2013). Physics: Principles with Applications (7th ed., pp.

431-442, 557-563). Pearson Education.

• Physics LibreTexts. (n.d.). Capacitors and capacitance.

Retrieved from https://phys.libretexts.org/Bookshelves/ Electricity_and_Magnetism/Capacitance

• Physics LibreTexts. (n.d.). Capacitors in series and parallel.

Retrieved from https://phys.libretexts.org/Bookshelves/ Electricity_and_Magnetism/Capacitors_in_Series_and_Parallel

Review Questions 3.1

1. State Coloumb’s law of electrostatics.

2. Sketch the electrostatic field due to two unlike point charges.

3. Two point charges 3 × 10⁻⁶C and 5 × 10⁻⁶C, are separated by a distance of 0.2 m. Calculate the magnitude of the electrostatic force between the two charges using Coulomb’s law. (Use k = 8.99 × 10⁹N⋅m²/C² as the electrostatic constant.)

4. Two point charges 2 × 10⁻⁶C and 4 × 10⁻⁶C exert an electrostatic force of 1.8 N on each other. Using Coulomb’s law, calculate the distance between the charges. (Use k = 8.99 × 10⁹N⋅m²/C² as the electrostatic constant.)

5. Three point charges are placed along a straight line. Charge q₁= 3 × 10⁻⁶ C is located at x = 0 m, charge q₂= -2 × 10⁻⁶C is located at x = 0.5 m, and charge q3 = 4 × 10⁻⁶C is located at x=1.0 m. Calculate the net electrostatic force acting on q₂due to q₁ and q₃. (Use k = 8.99 × 10⁹N⋅m²/C² as the electrostatic constant.)

6. Two identical spherical conductors, each with a charge of +5 μC, are initially placed 1 metre apart. When they are touched together, they share their charge equally and are then separated back to the original distance.

a. What is the new charge on each sphere after they are separated?

b. Calculate the electrostatic force between the two spheres after they have been re-separated. How does this force compare to the initial force when they each had a charge of +5 μC? Provide a detailed explanation of your reasoning.

Review Questions 3.2

1. Define the following terms:

a. Electric field strength

b. Electric potential

2. Using their formulae, derive SI units for electric field strength, and electric potential.

3. A charged particle is placed in an electric field with an electric field strength of 4.0 × 10³NC⁻¹and experiences a force of 0.8 N. Calculate the charge of the particle.

4. Two parallel conducting plates are separated by a distance of 0.08 m and have a potential difference of 550 V between them. Calculate the electric field strength between the plates.

5. Two parallel plates are separated by a distance of 0.05 m and connected to a 300 V power supply, creating a uniform electric field between them.

An electron (charge q = −1.6 × 10⁻¹⁹C and mass m = 9.1 × 10⁻³¹kg) is released from rest at the negative plate.

a. Calculate the electric field strength between the plates.

b. Determine the final speed of the electron when it reaches the positive plate, assuming no other forces are acting on it (hint: use conservation of energy principles to solve this problem)

Review Questions 3.3

1. What is the formula for calculating the capacitance of a capacitor?

2. What happens to the voltage across capacitors connected in parallel compared to those connected in series?

3. How does the capacitance of a parallel plate capacitor change if the distance between the plates is doubled?

4. You have three capacitors of values 2.0 μF, C₂=3.0 μF and C₃=6.0 μF connected in series. Calculate the total capacitance of the series combination.

5. You have three capacitors connected in series with capacitances of C₁=5.0 μF and C₂=10.0 μF, and the total capacitance of the series combination is given as Cₛ=2.0 μF; calculate the capacitance of the third capacitor, C₃.

Review Questions 3.4

1. What is the formula for calculating the energy stored in a capacitor?

2. How does the energy stored in a capacitor change if the voltage across it is doubled?

3. A capacitor with a capacitance of 100 μF is charged to a voltage of 12 V.

Calculate the energy stored in the capacitor.