Electricity, Magnetism & Circuits — Chapter Practice, Grade 12

repulsive

Like charges repel, so two positive charges push apart.

50000 N/C

E = F ÷ q = 0.1 ÷ (2 × 10⁻⁶) = 50,000 N/C.

0.6 N

F = kq₁q₂/r² = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶) ÷ (0.3)² = 0.054 ÷ 0.09 = 0.6 N.

copper

Copper is a metal with free-moving electrons, making it an excellent conductor — which is why wires are made of it.

0.9 N

F = kq₁q₂/r² = (9 × 10⁹)(3 × 10⁻⁶)(3 × 10⁻⁶) ÷ (0.3)² = 0.081 ÷ 0.09 = 0.9 N.

In a conductor, such as copper, some electrons are free to move through the material, so charge flows easily. In an insulator, such as rubber or glass, the electrons are held tightly to their atoms and cannot move freely, so charge does not flow. That is why wires use a copper conductor coated in a rubber insulator.

Distance is center-to-center under the point-charge model.

Away from the other charge: like charges repel, with equal and opposite interaction forces.

State what the supplied observations or conditions establish, and separate that from any assumption.

F=qE=6.0×10⁻⁴ N along the field for this positive charge.

Left, because the electron’s charge is negative and F=qE.

An electric field assigns force per positive test charge: E=F/q for a sufficiently small nonperturbing test. A positive source has outward field direction; a negative source has inward direction. A positive test charge feels force along E while a negative one feels force opposite E. Field units N/C differ from force units N. Field arrows are local direction indicators, not paths every released charge must follow.

Left, because the electron’s charge is negative and F=qE.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

F=qE gives+0.0010 N east for+2 μC and−0.0015 N, i.e.west, for−3 μC. The field is unchanged in the ideal nonperturbing-test model.

F=qE gives+0.0010 N east for+2 μC and−0.0015 N, i.e.west, for−3 μC.

The field is unchanged in the ideal nonperturbing-test model.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

4 A

I = Q ÷ t = 20 ÷ 5 = 4 A.

amperes

Current is the flow of charge per second, measured in amperes (amps).

12 V

V = W ÷ Q = 36 ÷ 3 = 12 V.

a short, thick wire

Resistance decreases with shorter length and greater thickness, so a short, thick wire resists the least.

20 C

Rearranging I = Q ÷ t gives Q = I × t = 2 × 10 = 20 C.

Voltage is like the water pressure from a pump that pushes the water along. Current is like the amount of water actually flowing past a point each second. Resistance is like a narrow section of pipe that restricts the flow. Higher pressure (voltage) drives more flow (current), while a tighter pipe (more resistance) reduces it.

Amperes are coulombs per second.

R=V/I=4 Ω.

State what the supplied observations or conditions establish, and separate that from any assumption.

No. Power P=VI also depends on voltage.

0.50 A under the same operating conditions.

Resistance relates voltage across a component to current through it; an ohmic model uses constant R so V=IR and the current–voltage graph is linear through the origin. Many real devices are nonohmic or change resistance as they heat. A curved graph does not violate energy conservation; it means a single constant-R approximation is inadequate. Read whether the vertical axis is current or voltage before interpreting the slope as R or 1/R.

0.50 A under the same operating conditions.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

I=9/12=0.75 A; P=VI=6.75 W; E=Pt=270 J. The stated constant resistance and voltage must persist over the interval.

I=9/12=0.75 A; P=VI=6.75 W; E=Pt=270 J.

The stated constant resistance and voltage must persist over the interval.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

10 V

V = IR = 2 × 5 = 10 V.

8 Ω

R = V ÷ I = 24 ÷ 3 = 8 Ω.

20 Ω

In series, resistances add: 4 + 6 + 10 = 20 Ω.

the same through all of them

A series circuit has one single path, so the identical current flows through every component.

6 Ω

1/R = 1/12 + 1/12 = 2/12 = 1/6, so R = 6 Ω. (For two equal resistors, the parallel total is always half of one.)

0.5 A

Rearranging P = IV gives I = P ÷ V = 60 ÷ 120 = 0.5 A.

Kirchhoff's loop balance provides a check.

R_eq=3 Ω; I_total=12/3=4 A, with 2 A per branch.

State what the supplied observations or conditions establish, and separate that from any assumption.

Each branch spans the same two electrical nodes, so it has the same potential difference.

Smaller for more than one finite positive branch in this ideal model.

An equivalent resistance gives the same voltage–total-current relation at a chosen pair of terminals as the ideal resistor network. Series resistances add; parallel conductances 1/R add. A 2 Ω equivalent for 6 Ω parallel with 3 Ω is smaller than either branch resistance because it offers more conducting paths. This does not mean either actual resistor changed its own resistance or that branch currents equal the total current.

Smaller for more than one finite positive branch in this ideal model.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Currents are 3 A and 1 A; total 4 A. Req=12/4=3 Ω and P=12×4=48 W. Branch powers 36 W and 12 W sum consistently.

Currents are 3 A and 1 A; total 4 A.

Req=12/4=3 Ω and P=12×4=48 W.

Branch powers 36 W and 12 W sum consistently.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

repel each other

Like magnetic poles repel, just as like charges do, so two north poles push apart.

passing an electric current through a coil of wire

A current flowing through a coil produces a magnetic field, forming an electromagnet that can be switched on and off.

the magnetic field through it is changing

Electromagnetic induction requires a changing magnetic field; a stationary field induces no current.

0.4 N

F = BIL = 0.2 × 4 × 0.5 = 0.4 N. The BIL form applies because current and field are perpendicular; otherwise include sinθ.

electrical energy into mechanical (motion) energy

A motor uses the force on a current-carrying coil in a magnetic field to turn electrical energy into rotational motion.

A generator spins a coil of wire inside a magnetic field (or spins a magnet near a coil). The constantly changing magnetic field through the coil induces a voltage by electromagnetic induction, driving a current. It is a motor in reverse because a motor takes in electricity to produce motion, while a generator takes in motion to produce electricity — the same parts, run the opposite way.

Zero force here does not mean no magnetic field.

F=2×3×0.5=3 N.

State what the supplied observations or conditions establish, and separate that from any assumption.

A changing flux produces an induced emf; current depends on the circuit being closed and its impedance.

Zero for a uniform field and flat loop, because θ=90° from the area normal.

Magnetic flux measures the field passing through an oriented surface. For a uniform field and flat area, ΦB=BAcosθ, where θ is between B and the area normal, not the surface itself. A loop can have nonzero field but zero net flux when its plane is parallel to B. Flux is measured in webers and differs from field strength in teslas. Changing field, loop area or orientation can change flux.

Zero for a uniform field and flat loop, because θ=90° from the area normal.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

F=|q|vB=0.012 N. Parallel motion gives sin 0=0 and zero magnetic-force magnitude in this model; electric forces would need a separate account.

F=|q|vB=0.012 N.

Parallel motion gives sin 0=0 and zero magnetic-force magnitude in this model; electric forces would need a separate account.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Force on the coil points toward the magnet; force on the magnet points toward the coil. They are equal/opposite and act on different bodies. Attraction was supplied as an assumption, not inferred from absent pole/current labels. An unchanged linked flux does not continuously induce emf; current also needs the appropriate conducting path and energy account.

Read source labels, scope, axes and intentional omissions.

Use the complete local explanation to draw a justified mechanism or calculate a supported quantity.

State what is observed, what is assumed and what evidence remains needed.

Response criteria

• Interpret the source labels, units and scope accurately.
• Explain the relevant mechanism with a labeled model or supported calculation.
• Identify a meaningful assumption, uncertainty or missing comparison.

Write the relationship first; substitute with consistent units.

12 J

qΔV=12.

Potential is scalar energy per charge; field is vector force per charge.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

0.012 C

Q=CV=.012.

Voltage grows during charging; average charging voltage is half final voltage for a linear capacitor.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Potential is per charge; energy change also depends on q.

Q=CV=4.0×10⁻⁵ C on either plate in magnitude.

State what the supplied observations or conditions establish, and separate that from any assumption.

½×4.0×10⁻⁶×100=2.0×10⁻⁴ J.

Q=CV=240 μC, with equal opposite signs on the two ideal plates.

A capacitor stores energy through separated opposite charges. Its capacitance C=Q/V uses the magnitude Q on either plate and the plate-to-plate potential difference, under a linear model. The entire ideal two-plate device can have zero net charge while each plate is charged. Farads are coulombs per volt. Geometry and dielectric material affect capacitance; doubling Q at fixed C doubles V rather than necessarily changing C itself.

Q=CV=240 μC, with equal opposite signs on the two ideal plates.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Q=400 μC; U=½×50×10⁻⁶×64=0.0016 J=1.6 mJ. At16 V, U=6.4 mJ, four times as much. Actual ratings must be separately respected.

Q=400 μC; U=½×50×10⁻⁶×64=0.0016 J=1.6 mJ.

At16 V, U=6.4 mJ, four times as much.

Actual ratings must be separately respected.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Write the relationship first; substitute with consistent units.

1.8 W

P=VI.

Energy; it is power multiplied by time.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

0.008 kWh

.004 kW×2h=.008 kWh.

Very low external resistance permits large current; internal resistance then dissipates substantial I²r heat.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Kilowatt-hour is energy, not power per hour.

1 kWh, equivalent to 3.6×10⁶ J.

State what the supplied observations or conditions establish, and separate that from any assumption.

The 10 W lamp uses less energy per equal operating time; broader impact also requires lifetime and other factors.

3 A and 36 W at unchanged 12 V, assuming the source maintains that voltage.

For ideal parallel loads sharing 12 V, source current is the sum of branch currents. Here 2+3+1=6 A and source output power is 72 W. A real source and conductors have operating limits and losses, so an ideal current sum is only part of a design comparison. Consider voltage regulation, continuous versus peak demand and the stated component ratings. Use supplied-data design here rather than modifying household wiring.

3 A and 36 W at unchanged 12 V, assuming the source maintains that voltage.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Saving=(0.015−0.009)×2500=15 kWh=5.40×10⁷ J/year. Production, transport, replacement lifespan or disposal is outside this operational-only result.

Saving=(0.015−0.009)×2500=15 kWh=5.40×10⁷ J/year.

Production, transport, replacement lifespan or disposal is outside this operational-only result.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.