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Unit notes and practice questions

Grade 10 Physics unit 4: Static and Current Electricity

Everything here is for unit 4 of the Grade 10 Physics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

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6
Cards
10
Practice questions on unit 4
82–131
The textbook pages for unit 4

What unit 4 covers

  1. 1.Charges in Nature
  2. 2.Methods of Charging a Body
  3. 3.The Electroscope
  4. 4.Electrical Discharge
  5. 5.Coulomb's Law of Electrostatics
  6. 6.The Electric Field
  7. 7.Electric circuits
  8. 8.Current, Voltage and Ohm's law
  9. 9.Combination of Resistors in a Circuit
  10. 10.Voltmeter and Ammeter Connection in a Circuit
  11. 11.Electrical Safety in General and Local context
  12. 12.Electric Projects

Short notes for unit 4

The cards filed under this chapter on the Grade 10 Physics page. Open one at a time.

F=kq1q2r2F = k \frac{q_1 q_2}{r^2}
FF
electric forceN\text{N}
kk
Coulomb constantN m2/C2\text{N}\,\text{m}^2/\text{C}^2
q1q_1
electric charge of first particleC\text{C}
q2q_2
electric charge of second particleC\text{C}
rr
distance between chargesm\text{m}

When you use it

Use to calculate the electrostatic force of attraction or repulsion between two point charges.

Watch out

Remember to square the distance in the denominator. A negative result indicates attraction and a positive result indicates repulsion.

Drafted from Grade 10 Physics, pages 82-131, then checked twice before it went up

E=kqr2E = k \frac{q}{r^2}
EE
electric field strengthN/C\text{N}/\text{C}
kk
Coulomb constantN m2/C2\text{N}\,\text{m}^2/\text{C}^2
qq
source electric chargeC\text{C}
rr
distance from the chargem\text{m}

When you use it

Use to calculate the electric field strength at a specific distance from a point charge.

Watch out

Distance r must be measured in metres and squared in the denominator.

Drafted from Grade 10 Physics, pages 82-131, then checked twice before it went up

When you use it

Use when configuring meters to measure current or voltage in an electric circuit.

Watch out

Never connect an ammeter in parallel or a voltmeter in series. An ammeter in parallel draws excess current and gets damaged, while a voltmeter in series prevents current from flowing due to its high resistance.

Drafted from Grade 10 Physics, pages 82-131, then checked twice before it went up

ΔV=IR\Delta V = I R
ΔV\Delta V
potential differenceV\text{V}
II
electric currentA\text{A}
RR
electrical resistanceΩ\Omega

When you use it

Use to relate potential difference, electric current, and resistance across an electrical component.

Watch out

Ohm law applies directly to Ohmic conductors where physical conditions such as temperature remain constant.

Drafted from Grade 10 Physics, pages 82-131, then checked twice before it went up

ConnectionEquivalent resistanceCurrentPotential difference
SeriesReq=R1+R2+⋯+RnR_{eq} = R_1 + R_2 + \dots + R_nI=I1=I2=…I = I_1 = I_2 = \dotsΔV=ΔV1+ΔV2+…\Delta V = \Delta V_1 + \Delta V_2 + \dots
Parallel1Req=1R1+1R2+⋯+1Rn\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}I=I1+I2+…I = I_1 + I_2 + \dotsΔV=ΔV1=ΔV2=…\Delta V = \Delta V_1 = \Delta V_2 = \dots

When you use it

Use when finding the equivalent resistance, branch currents, or voltages in series and parallel resistor circuits.

Watch out

For parallel combinations, solve for 1/R_eq first and take the reciprocal at the end to get R_eq.

Drafted from Grade 10 Physics, pages 82-131, then checked twice before it went up

Part of the circuitIn normal useWhen a fault happens
Live wireBrings current to the appliance at about 220 V with respect to the earth.If it touches the metal body, the body goes live until the supply is cut off.
Neutral wireReturns the same current to the supply and sits near 0 V.It carries the working current, so it can never stand in for the earth wire.
Earth wireCarries no current at all and holds the metal body at earth potential.Gives the fault current a low-resistance path to the ground, so a large current flows.
Fuse or MCBSits in the live wire and never in the neutral.The large fault current blows the fuse or trips the MCB, and the appliance is disconnected.
Two-pin plugGives a metal-bodied appliance no earth path.A fault can leave the casing live. A double-insulated appliance with a plastic casing is built for two pins on purpose.
The socket in an Ethiopian wallEuropean round-pin, 220 V at 50 Hz, with the earth contact often made by clips on the sides.The three rectangular pins drawn on most revision sheets are the British pattern, where the earth pin is longer and thicker so it makes contact first and breaks last.

When you use it

Use when a question asks why a metal-bodied appliance such as a refrigerator, an electric mitad or a water heater needs a third wire. The earth wire protects the person touching it: it gives fault current a low-resistance path, so the fuse blows or the MCB trips before anyone is hurt.

Watch out

Students treat the earth wire and the neutral wire as the same thing because both measure near 0 V, and some join the earth pin straight to the neutral to save a wire. They are not the same. In normal use the neutral carries the full working current while the earth carries none, so if a neutral used as an earth breaks anywhere upstream, the metal body is joined through the motor to the live wire and sits at 220 V. The other half of the same mistake is fitting the fuse in the neutral: it still blows, but the appliance stays connected to the live wire and is dangerous to touch. The rule is short: the fuse goes in the live wire, the earth goes to the metal body, and the neutral carries the current back to the supply.

Practice questions on unit 4

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

According to the law of conservation of charge, what happens to electric charges in an isolated system?

  1. A.They can be created from other forms of energy.
  2. B.They can neither be created nor destroyed, only transferred.
  3. C.They are completely destroyed when opposite charges touch.
  4. D.They continuously increase in number over time.
Show the answer and why

Answer: B They can neither be created nor destroyed, only transferred.

Why

The law of conservation of charge states that electric charges can neither be created nor destroyed, but can only be transferred from one material to another.

Question 2

A student in Hawassa uses a gold-leaf electroscope in the laboratory. What is the primary function of an electroscope?

  1. A.To produce a continuous electric current
  2. B.To measure the electrical resistance of a circuit
  3. C.To detect the sign and amount of charge on a body
  4. D.To convert electrical energy into mechanical energy
Show the answer and why

Answer: C To detect the sign and amount of charge on a body

Why

An electroscope is a device specifically used to detect the sign and amount of electric charge on a body.

Question 3

What is the potential difference between two points when 1 Joule of work is done to move a charge of 1 Coulomb from one point to the other?

  1. A.1 Ampere
  2. B.1 Volt
  3. C.1 Ohm
  4. D.1 Watt
Show the answer and why

Answer: B 1 Volt

Why

By definition, one volt is the potential difference between two points in a conductor when 1 Joule of work is done to move a charge of 1 Coulomb from one point to the other (ΔV = W / Q).

Question 4

Why must an ammeter be connected in series with a resistor in an electric circuit?

  1. A.Because components in series experience the same potential difference.
  2. B.Because components in series have the same electric current passing through them.
  3. C.Because connecting it in series increases circuit resistance to infinity.
  4. D.Because it prevents current from flowing through the rest of the circuit.
Show the answer and why

Answer: B Because components in series have the same electric current passing through them.

Why

An ammeter is connected in series because objects connected in series have the exact same current flowing through them.

Question 5

According to Coulomb's law, what happens to the electrostatic force between two point charges if the distance between them is doubled?

  1. A.The force becomes 4 times smaller.
  2. B.The force is doubled.
  3. C.The force becomes 2 times smaller.
  4. D.The force becomes 4 times larger.
Show the answer and why

Answer: A The force becomes 4 times smaller.

Why

Coulomb's law states that the electrostatic force is inversely proportional to the square of the distance (F ∝ 1/r2). Doubling the distance increases the denominator by 22 = 4, making the force 4 times smaller.

Question 6

What is the magnitude of the electric field at a point 2.00 m away from a point charge of 5.00 mC (5.00 × 10-3 C)? (Use k = 9 × 109 N m2/C2)

  1. A.2.25 &times; 10<sup>6</sup> N/C
  2. B.4.50 &times; 10<sup>6</sup> N/C
  3. C.11.25 &times; 10<sup>6</sup> N/C
  4. D.22.50 &times; 10<sup>6</sup> N/C
Show the answer and why

Answer: C 11.25 &times; 10<sup>6</sup> N/C

Why

Using the formula E = k · q / r2:
E = (9 × 109 N m2/C2 × 5.00 × 10-3 C) / (2.00 m)2 = 45 × 106 / 4 = 11.25 × 106 N/C.

Question 7

Two copper wires in a workshop in Addis Ababa have the same length. Wire A has twice the cross-sectional area of Wire B. How does the resistance of Wire A compare to Wire B?

  1. A.Wire A has twice the resistance of Wire B.
  2. B.Wire A has four times the resistance of Wire B.
  3. C.Wire A has half the resistance of Wire B.
  4. D.Wire A has the same resistance as Wire B.
Show the answer and why

Answer: C Wire A has half the resistance of Wire B.

Why

The resistance of a conductor is inversely proportional to its cross-sectional area (R = ρl / A). Therefore, doubling the area reduces the resistance by half.

Question 8

Three resistors of 4.0 Ω, 8.0 Ω, and 12.0 Ω are connected in series across a 24.0 V power source. What is the electric current flowing through the circuit?

  1. A.0.5 A
  2. B.1.0 A
  3. C.2.0 A
  4. D.3.0 A
Show the answer and why

Answer: B 1.0 A

Why

For resistors in series, equivalent resistance is Req = 4.0 Ω + 8.0 Ω + 12.0 Ω = 24.0 Ω. Using Ohm's law, I = ΔV / Req = 24.0 V / 24.0 Ω = 1.0 A.

Question 9

Two charges, q1 = +2 × 10-6 C and q2 = -4 × 10-3 C, are separated by 0.3 m. What is the magnitude and nature of the electric force between them? (Use k = 9 × 109 N m2/C2)

  1. A.800 N, attractive
  2. B.800 N, repulsive
  3. C.240 N, attractive
  4. D.240 N, repulsive
Show the answer and why

Answer: A 800 N, attractive

Why

Using Coulomb's law: F = k · |q1 · q2| / r2 = (9 × 109 × 2 × 10-6 × 4 × 10-3) / (0.3)2 = 72 / 0.09 = 800 N. Since the two charges have opposite signs, they attract each other.

Question 10

A piece of wire is cut into 5 identical pieces. When connected in parallel, their equivalent resistance is 2 Ω. What was the resistance of the original wire before it was cut?

  1. A.10 &Omega;
  2. B.25 &Omega;
  3. C.50 &Omega;
  4. D.100 &Omega;
Show the answer and why

Answer: C 50 &Omega;

Why

Let each piece have resistance R. In parallel, 1/Req = 5/R, which gives R = 5 × 2 Ω = 10 Ω for each piece. Originally, the 5 pieces were in series, so the total original resistance was 5 × 10 Ω = 50 Ω.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 4

Unit 4 runs from page 82 to page 131 of the Ministry of Education edition.

Ministry of Education edition, English, 212 pages

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