Subject: 

PHYSICS

[mediator_tech]

Term:

FIRST TERM

Week:

WEEK 2

Class:

SS 3

Topic:

ELECTRIC FIELDS

[mediator_tech]

Previous lesson: 

The pupils have previous knowledge of

Gravitational Force between Two Masses (Newton’s Law of Universal Gravitation)

that was taught as a topic in the previous lesson

 

Behavioural objectives:

At the end of the lesson, the learners will be able to

1. Mention an instrument used for comparing the emf of cells.
2. Write down the formulae for net resistance and emf in a circuit.
3. Define conductivity.
4. Write down the relationship between conductivity and resistivity.

 

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

 

[mediator_tech]

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

[mediator_tech]

Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks

 

Content:

 

 

ELECTRIC FIELDS

CONTENT

1. Production of Continuous Charges
2. Types of Cells
3. Electric Circuit Series and Parallel Arrangement of Cells and Resistors
4. Shunts and Multipliers
5. Resistivity and Conductivity
6. The Wheatstone Bridge
7. The Metre Bridge
8. The Potentiometer
9. Merits of the Potentiometer Over the Voltmeter
10. Electric Force between Two Points Charges (Coulomb’s Law of Electrostatic Force)
11. Electric Field Intensity or Strength ε
12. Electric Potential V_E

 

Production of Continuous Charges

Continuous charges are also called current (or moving electrons). They are produced by a device called ‘a cell’.

In some cells, a chemical reaction occurs between two different metals and a liquid electrolyte that generates a continuous charge. The most common types of cells used in science experiments are the Daniell cell and the Leclanche cell. These cells contain two different electrodes (anode and cathode), which form an electrolytic solution in an outer container called the ‘cell’. The anode is made of zinc and the cathode is made of copper or carbon.

Types of Cells

There are two types of cells. Namely:

A. Primary cells

B. Secondary cells.

A. Primary cells:

These are cells that cannot be recharged once they run down. E.g. torch batteries, calculator batteries, leclanche cell, Daniel cell. The components of the cell are shown below:

1. Anode: The anode is the negative terminal of the cell, and it is made of zinc.

2. Cathode: The cathode is the positive terminal of the cell, and it is made of copper or carbon.

3. Electrolyte solution: This liquid electrolyte contains ions that conduct electricity.

 

However, primary cells have two defects. They are:

1. Polarization: This is a defect that results from the production of hydrogen bubbles around the copper plate of the cell. This will set up a back emf that reduces the emf in the external. As the cell works, hydrogen bubbles are released. These bubbles insulate the positive electrode and then slow it down. It could be minimized by brushing the plate or by using a depolarizer called manganese (iv) oxide (MnO2), Potassium tetraoxomanganese(IV) KMnO¬, Potassium Heptaoxodichromate (VI) K₂Cr₂O₇… to oxidize the hydrogen bubbles to form water.
2. Local Action: This defect results from not using a pure zinc plate which makes it to wear It is reduced by amalgamation i.e cleaning the zinc plate with H₂SO₄ and then rubbing it with mercury.

B. Secondary cells:

These are cells that could be recharged when they run down using a suitable battery charger. The two major types are lead acid accumulator and the nickel cadmium cell. E.g car batteries, motorcycle batteries, cell phone batteries and all other rechargeable batteries.

These are cells that can be recharged once they run down, by applying an electric current to them with a battery charger or other device. Common examples of secondary cells include lead-acid batteries and nickel-cadmium batteries.

Electric Circuit Series and Parallel Arrangement of Cells and Resistors

(a) Cell Arrangement:

(i) Cells in Series:

The total emf is the sum of all the emfs.

 

E_T = E1 + E2 + E3

(ii) Cells in Parallel:

All emfs are equal.

E_T = E1 or E2 or E3, if E1 = E2 = E3, i.e, all emfs are equal.

E_T = E3, (emf with the highest value), if E3 > E2 > E1

(b) Resistors Arrangement:

(i) Resistors in Series:

 

V_T = V₁ + V₂ + V₃       

(Circuital components in series have the same I but different V)

But V = IR

∴ IRT = IR1 + IR2 + IR3

IRT = I(R1 + R2 + R3)

Dividing through by I, we have R_T = (R₁ + R₂ +R₃)

(ii) Resistors in Parallel:

 

IT=I1+I2+I3

But I=VR

∴VRT=VR1+VR2+VR3∴V(1RT)=V(1R1+1R2+1R3)

Dividing through by V, we have:

(1RT)=(1R1+1R2+1R3)=R1R2+R1R3+R2R3R1R2R3

Examples:

1. If four 3 resistors are connected a) in series b) in parallel, find their net resistance in each case.

Solution:

(a) In series,

RT=R1+R2+R3+R4=3+3+3+3=12Ω

(b) In parallel,

(1RT)=(1R1+1R2+1R3+1R4)=(1RT)=(13+13+13+13)=43RT1=34RT=0.75Ω

2. Three cells of value 2V, 2V and 3V are connected (a) in series (b) in parallel. Find their total emf in each case.

Solution:

(a) In series,

ET=E1+E2+E3=2+2+3=7V

(b) In parallel, since 3V has the highest emf among the three cells, E_T = 3V

EVALUATION

1. State the formula for finding the net emf and resistance when these components are in a) series b) parallel.
2. When two cells of value 2V and 3V are in parallel, what is their total emf?
3. ting or Short-circuiting

Shunts and Multipliers

1. Shunt:

A shunt is a low resistance usually used to convert a galvanometer to an ammeter. It is usually connected in parallel with the galvanometer.

I – Current to be measured

I_g – Current through the galvanometer

R_s – Resistance of the shunt

R_g – Resistance of galvanometer

Note: The voltage across the shunt is the same as that across galvanometer.

Vg=VsIgRg=(I−Ig)Rg

Example:

A galvanometer gives a full scale deflection when a current of 10mA flows through it. How would you convert it to an ammeter capable of reading 3Aif the resistance of the galvanometer is 5Ω?

Solution:

To do this, we connect a resistor of low resistance called shunt in parallel with the galvanometer. The value of such shunt is calculated below:

10mA=101000=0.01A

p.d across galvanometer = p.d across shunt

Igrg=IsRsRs=IgrgIs=0.01×52.99=0.0167Ω

Thus, a shunt of 0.0167Ω is needed.

2. Multiplier:

A multiplier is a high resistance usually used to convert a galvanometer to a voltmeter. It is usually connected in series with the galvanometer.

Example:

To convert a galvanometer to a voltmeter ,we need a multiplier Rso that the galvanometer could read 20V at full scale deflection of 10mA. Find the value of R if the internal resistance of the galvanometer is 6Ω. Solution:

 

r=6Ω20v=V1+V2∴20=0.01r+0.0RR=1994Ω

 

Resistivity and Conductivity

Experiment shows that the resistance of a metallic conductor is directly proportional to the length of the wire and inversely proportional to the cross-sectional area.

i.e, R∝LA∴R=ρLA

Hence, ρ=RAL

Where ρ the constant of the equation called the resistivity, expressed in ohm-meter (Ωm)

Resistivity of a wire is defined as the product of the wire’s resistance and its cross sectional area, divided by the length of that wire. It can also be defined as the resistance of unit length of material of unit cross sectional area. It is the ability of a substance to restrict heat.

If a wire has a diameter d, then

A=πr2=πd24∴R=4ρLπd2andρ=πRd24L

Conductivity on the other hand, is a measurement of the ability of a substance to conduct heat. It is the reciprocal of resistivity and denoted by σ

Hence, σ=1ρσ=LRA=4LπRd2

Example:

A2m resistance wire, area of cross-section 0.50mm², has a resistance of 2.20Ω. Calculate

(a) the resistivity of the wire.

(b) the length of a wire which, when connected in parallel with the 2m wire will give a resistance of 2.0 Ω.

Solution:

(a) ρ=RALρ=2.2×0.5×10−62ρ=5.5×10−7Ωm

EVALUATION

1. Define resistivity.
2. What is the relationship between resistivity and conductivity?
3. What is a shunt?
4. -Franz law states that the resistivity of a metal wire is directly proportional to its absolute temperature.
5. Explain this statement.
6. What are the limitations of Wiedemann–Franz law?
7. Resistivity is a measure of how easily an electrical current flows through a material.

 

The Wheatstone Bridge

The Wheatstone bridge is one of the most accurate methods of measuring resistance. It consists of four resistors of resistance R₁, R₂, R₃ and R₄ connected as shown below.

 

When no current flows through the galvanometer, the bridge is said to be balanced.

Therefore, p.d across R₁ = p.d across R₃

And p.d across R₂ = p.d across R₄ 

Hence, I1R1=I2R3 ——–(1) and

I1R2=I2R4 ——–(2)

Dividing (1) by (2), we have,

R1R2=R3R4

Where R₁ is the unknown resistance, R₂ is a fixed resistance of known value, R₃ & R₄ are variable resistances of known values.

 

The Metre Bridge

It consists of a straight uniform resistance wire AB of length 1m stretched along a metre rule. The unknown resistor X is placed at the left side, while the known R is placed at the right side.

 

When the circuit is closed, a point is located along the resistance wire with the jockey when G reads zero.

At this point, X ∞ l₁ and R ∞ l₂

∴ X = kl₁ and R = kl₂

Since k is the same,

∴XI1=RI2∴X=RI1I2

Precautions:

1. The value of the known resistance R should be chosen so that the balance point comes between 30cm and 70cm.
2. The battery key should always be depressed before the G contact is made on the bridge wire.
3. Avery small current should be passed through the galvanometer to avoid damage.
4. In other not to destroyed the uniformity of the wire, the jockey should touch the wireentiometer
5. It is used to measure the internal resistance of a cell, the emf and internal resistance of two cells in series or parallel, p.d. across two resistances in series and emf of a single cell.

1. The potentiometer method is based on the fact that the potential difference between points A and C in a circuit with a uniform conductor AB at any point along the wire, produced by a unit current I flowing from A to B through AB depends only on LC.

 

The Potentiometer

A potentiometer consists of uniform resistance wire AB of length 100cm through which a source of emf maintains a steady current.

Since the wire is uniform and current constant,

V∝R∝IV=kL

 

When a potentiometer is used to compare the emfs of two cells, at balance point when G reads zero,

E1E2=I1I2

 

Merits of the Potentiometer Over the Voltmeter

1. The potentiometer is a more accurate in measuring voltage than the voltmeter.
2. There is no zero scale error as that associated with pointer instrument like voltmeter.
3. It passes no current at the time current is being taken. power, one of the basic concepts in physics, is a measure of how much work an electrical system can do. Typically, power is measured in joules per second or watts (J/s), and it is related to voltage (voltage x current) and resistance by the following equation:
4. P=VIR
5. where P is electric Shock
6. An electric shock occurs when an object with a high voltage potential has a conductive path to the body. In most cases, this is caused by faulty wiring or improper use of electrical equipment. Electric shocks can cause serious injuries and even death if they are not treated immediately. Some common symptoms include loss of muscle control, burning sensation, nausea, Shock Prevention and Treatment
7. There are a number of steps that you can take to avoid electric shock, including:
8. · Always use properly rated electrical cords and equipment. Never use extension cords or adapters unless they are designed for this purpose.
9. · Avoid using electrical equipment in wet conditions. Use caution when using power tools near water sources or in damp environments

 

Electric Force between Two Points Charges (Coulomb’s Law of Electrostatic Force)

Like charges repel and unlike charges attract. Based on this, Coulomb’s law of electrostatic force came into being. The law states that, ‘The electrostatic force of attraction or repulsion F_C between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their distance apart.’

Mathematically,

FC∝Q1Q2r2FC=kQ1Q2r2

Where k=14πϵ0

Hence, FC=Q1Q24πϵ0r2

Where ϵ0=8.85×10−12F/m

and k=9×109Nm2/C2

 

Electric Field Intensity or Strength ε

This is defined as the electric force experienced per unit charge. It is a vector quantity and is expressed in N/C.

Mathematically,

E=FCQ∴E=Q4πϵ0r2

 

Electric Potential V_E

The electric potential V_E at a point in an electric field is defined as the work done in taking a unit charge from infinity to that point. It is a scalar quantity and is expressed in volt.

Mathematically,

VE=Er

But E=Q4πϵ0r2VE=(Q4πϵ0r2)rVE=Q4πϵ0r

 

Worked Examples

Example 1:

Calculate the electrostatic force that exists between two charges of values 5.3μC and -3.2μC placed at 6.0 × 10^-4m. (k = 9 × 10⁹Nm²/C²)

Solution:

FC=kQ1Q2r2FC=9×109×5.3×10−6×3.2×10−6(6.0×10−4)2=4.24×105N

Example 2:

Determine the electric potential due to charge 5.3μC placed at that point.

Solution:

VE=Q4πϵ0r=5.3×10−6×9×1096×10−4=79,500KV

Example 3:

Three charges +10C, -20C and 16C are distributed as shown in the figure below. Find the resultant force acting on the 10C charge.

Solution:

Let the Force of attraction between +10C and – 20C be

F1=kQ1Q2r2F1=9×109×10×2032=2×1011N(k=14πϵ0r29×109)

Let the Force of attraction between +10C and – 16C be

F2=kQ1Q2r2F2=9×109×10×162×2=3.6×1011N(3.6×1011)2

Net force = (F21+F22)−−−−−−−−√=(2×1011)+(3.6×1011)2−−−−−−−−−−−−−−−−−−−−√=4.2×1011NTanθ=F1F2=2×10113.6×1011=0.555θ=tan−10.555=32.28o

The net force is 4.12 × 10¹¹N in a direction 32.28⁰ to the line joining the +10C and – 16C charges.

Example 4:

Two similar but opposite point charges –q and +q each of magnitude 5.0 x 10 ^–8 C are separated by a distance of 8.0cm in a vacuum as shown below.

Calculate the magnitude and direction of the resultant electric field intensity Eat the point p.

k=14πϵ0r29×109Nm2C−2

Solution:

At p the direction of the field vector is towards the left since the positive charge at p is attracted towards the –q charge.

E−=−q4πϵ0r2=5.0×10−8×9×109(0.05)−2E−=1.8×105NC−1

The field vector due to +q charge is also directed towards the left since the positive charge at p is repelled by +q

E+=+q4πϵ0r2=5.0×10−8×9×109(0.03)−2E+=5×105NC−1

Resultant electric field is E=E−+E+

E=(1.8×105+5.0×105)NC−1

The field is directed towards the left or towards –q charge.

 

EVALUATION

1. State two advantages of potentiometer over voltmeter.
2. Mention two precautions to be ensured in using the metre bridge.
3. State Coulomb’s law.
4. Define electric potential.
5. One advantage of using a potentiometer to measure voltage is that it is a more sensitive measurement than a voltmeter. This makes it an ideal tool for high-precision measurements, such as those required in laboratory settings or scientific research.
6. Another advantage of using a potentiometer is that it provides accurate readings even atATION
7. In order to ensure accurate and reliable measurements using a meter bridge, it is important to take certain precautions into account. These include ensuring that the two sides of the bridge are balanced prior to taking any readings, as well as being careful not to touch the conductors or wires while making measurements.

1. Coulomb’s law states that

GENERAL EVALUATION

1. Mention an instrument used for comparing the emf of cells.
2. Write down the formulae for net resistance and emf in a circuit.
3. Define conductivity.
4. Write down the relationship between conductivity and resistivity.

1. One common instrument used for comparing the emf of cells is an ohmmeter. This device measures resistance and allows users to directly compare how much electrical current will flow through two different cells.
2. the net resistance is calculated using the formula R=Σr1+r2+r3+…, where r1, r2, and r3 represent the resistances of each individual component in the circuit
3. circuit is dependent on a number of factors, including the total resistance, the source voltage, and any external resistances in the circuit. The general relationship between these quantities can be expressed as E=IR, where I represents current and R represents resistance.
4. , or the ability of a material to allow electricity to flow through it, is determined by a number of factors, including its resistivity and temperature. Generally speaking, higher temperatures will result in lower conductivity, as thermal motion can interfere with the movement of electrons in materials. Additionally, substances with larger atomic radii generally have higher conductivities, as
5. Conductivity refers to a material’s ability to conduct electricity, while resistivity is a circuit is related to the total resistance of the circuit, as well as the source voltage. Specifically, emf can be expressed in terms of Ohm’s law as E=IR, where E represents emf, I represents current, and R represents resistance. Thus, increasing the total resistance or source voltage will result in a higher emf for a circuit is determined by many factors, including the total resistance and source voltage within that circuit. In terms of Ohm’s law, emf can be expressed as E=IR, where I represents current and R represents resistance. Thus, increasing the resistances or source voltage in a circuit will result in a higher emf reading. Furthermore, materials with circuit is determined by a number of factors, including the total circuit resistance and source voltage. In terms of Ohm’s law, the emf can be expressed as E=IR, where I represents current and R represents resistance. Thus, increasing either of these quantities will result in a higher emf reading.
6. The emf of a cell can also be expressed mathematically as E = IR + Ir, where I represents
7. In a circuit, the net resistance is equal to the reciprocal of the total conductance in that circuit. In other words, it is given