Explain Energy Transfer Via Fields - Grade 11 Physics

An electron begins nearly at rest and moves through an electric field from a location at $0 V$ to a location at $+120 V$ . Ignore all other forces. What happens to the electron's energy?
Answer: Its electric potential energy decreases by $120 eV$ , and its kinetic energy increases by $120 eV$ . Whenever you see a question about a charged particle moving through a potential difference, your first instinct should be to ask: what sign is the charge, and which way does the electric force actually push it? For an electron (charge $q = -e$ ), the electric potential energy is given by $U = qV$ . Moving from $0\text{ V}$ to $+120\text{ V}$ , the change in potential energy is: $\Delta U = q \Delta V = (-e)(+120\text{ V}) = -120\text{ eV}$ The potential energy decreases by $120\text{ eV}$ . By conservation of energy (with no other forces doing work), that lost potential energy must convert entirely into kinetic energy, so $\Delta KE = +120\text{ eV}$ . This confirms D is correct — and it makes physical sense: the electric force on a negative charge points opposite the field, meaning it pushes the electron toward higher potential, accelerating it. A is wrong on two counts: while it correctly identifies that the field decelerates a negative charge moving toward higher potential... wait, actually it doesn't — the force on a negative charge toward higher potential is in the same direction as motion, so the electron speeds up, not slows down. The energy accounting is also inverted. B gets the kinetic energy increase right but claims potential energy also increases — that violates conservation of energy. Both quantities cannot increase simultaneously without an external energy source. C introduces a fictional idea that energy stays "in the field" rather than transferring to the electron — conservation of energy requires the lost potential energy goes directly into the electron's kinetic energy. Study tip: Always apply $\Delta U = q\Delta V$ using the actual charge (negative for electrons). The sign does all the work for you.
A positive charge of $2.0 μC$ is released from rest at a point where the electric potential is $40 V$ . It moves only under the electric force to a point where the potential is $10 V$ . Which statement correctly describes the transfer of energy?
Answer: The electric potential energy decreases by $60 μJ$ , and the electric field transfers $60 μJ$ to the charge's kinetic energy. When a charge moves through an electric potential difference, energy is conserved between two forms: electric potential energy (EPE) and kinetic energy. The key formula here is: $\Delta EPE = q \Delta V = q(V_f - V_i)$ Here, $q = 2.0 \times 10^{-6}\ C$ , $V_i = 40\ V$ , and $V_f = 10\ V$ , so: $\Delta EPE = (2.0 \times 10^{-6})(10 - 40) = (2.0 \times 10^{-6})(-30) = -60\ \mu J$ The EPE decreases by $60\ \mu J$ . Since the charge starts from rest and only the electric force acts, energy conservation requires that this lost potential energy becomes kinetic energy — the electric field does $60\ \mu J$ of work on the charge. This confirms A is correct. B is wrong on two counts: EPE doesn't increase (the charge moves from high to low potential, which for a positive charge means losing EPE), and kinetic energy is gained, not transferred back to the field. C uses the right reasoning but makes a calculation error — it likely confuses the potential difference ( $30\ V$ ) with the initial or final potential value, arriving at the wrong $\Delta EPE$ of $20\ \mu J$ . D reflects a fundamental misconception. A change in electric potential absolutely affects a charge — that's precisely the mechanism by which the electric force accelerates it. Study tip: For positive charges, moving from high to low potential always means losing EPE and gaining kinetic energy — think of it like a ball rolling downhill. For negative charges, the relationship flips.
An ideal transformer has a primary voltage of $120 V$ and a primary current of $0.50 A$ . Its secondary voltage is $24 V$ . Which statement correctly describes the secondary current and the transfer of energy?
Answer: The secondary current is $2.5 A$ , and energy is transferred between the coils by changing electromagnetic fields. Whenever you see a transformer question, two tools should come to mind immediately: the turns-ratio equation for voltage and the power conservation equation for current (since an ideal transformer wastes no energy). For an ideal transformer, input power equals output power: $P_{primary} = P_{secondary}$ , which means $V_p I_p = V_s I_s$ . Plugging in the given values: $120 \times 0.50 = 24 \times I_s$ , so $I_s = \frac{60}{24} = 2.5 \text{ A}$ . Notice that because the secondary voltage is lower (a step-down transformer), the secondary current is higher — the transformer trades voltage for current while conserving power. Energy is transferred between the coils through changing electromagnetic fields (mutual induction): the alternating current in the primary creates a changing magnetic flux through the core, which induces a voltage in the secondary. No electrons cross between coils. Choice A gets the current wrong (0.10 A would require the secondary voltage to be higher than the primary), and electrons absolutely do not travel through the transformer core — that's not how electromagnetic induction works. Choice C has the correct current but contains a dangerous misconception: transformers don't produce energy. The lower secondary voltage doesn't generate anything; it simply reflects the turns ratio. Choice D is wrong on both counts. The current is not 0.50 A, and energy is not trapped in the primary — the whole purpose of a transformer is to transfer energy, which it does via electromagnetic fields even though the coils are electrically isolated. Your study tip: always use $V_p I_p = V_s I_s$ for current in ideal transformers, and remember — step-down voltage means step-up current.
A spacecraft approaches a planet moving around its star, passes behind the planet relative to the planet's orbital motion, and departs with greater speed relative to the star. The spacecraft's engines remain off throughout the encounter. Which statement best explains the spacecraft's energy gain?
Answer: The moving planet transfers a small amount of its orbital energy to the spacecraft through their changing gravitational field interaction, so the spacecraft gains speed while the planet loses a negligibly small amount of orbital energy. When you see a question about gravitational assists (also called "slingshot maneuvers"), the key concept to keep in mind is energy conservation in a two-body system — specifically, how energy can be transferred between objects through gravity when one of them is already moving. In a gravitational assist, the planet isn't stationary — it's orbiting the star at high speed. When the spacecraft passes behind the planet relative to its orbital motion, the planet's gravity pulls the spacecraft forward, accelerating it. Simultaneously, the spacecraft's gravity pulls back on the planet, causing the planet to lose a tiny (negligibly small) amount of orbital energy. Total energy is conserved across the system; it's simply redistributed. This is exactly what A describes, making it the correct answer. B is wrong because it treats the planet's gravitational field as stationary, which misses the entire point. A static gravitational field cannot add net energy — a spacecraft falling in and escaping symmetrically would return to its original speed. The planet's motion is what drives the energy transfer. C contains a common misconception: that falling toward a planet and escaping it yields a net energy gain from the field alone. In reality, a spacecraft interacting with a truly stationary planet would return to exactly its original speed upon escape — no permanent gain occurs. The gain requires the planet's orbital motion. D is wrong because the star's gravity isn't what powers the flyby; the energy comes from the planet's orbital motion, not from the star acting directly on the spacecraft. A good study tip: whenever gravitational assists appear, ask yourself, "what is moving here, and who loses energy?" The answer will always point to the planet's orbital motion as the energy source.
An ideal transformer's coils are electrically isolated. What transfers energy between them?
Answer: Changing magnetic field In an ideal transformer, the primary coil creates a changing magnetic field, and that field links to the secondary coil, inducing a voltage there. That magnetic coupling is what transfers energy across the electrically isolated coils. The tempting wrong answer is electric field in the gaps: the coils are isolated, so the energy isn't carried by moving charge or an electric field between them.
Electromagnetic radiation with an intensity of $800 W/m^2$ strikes a detector of area $0.020 m^2$ for $30 s$ . The detector converts $25%$ of the incident energy into electrical energy. How much electrical energy is produced, and how did the energy reach the detector?
Answer: $120 J$ is produced; electromagnetic fields carried the incident energy through space to the detector. When you see a question combining electromagnetic radiation with energy conversion, tackle it in two parts: calculate the total incident energy, then apply the efficiency factor — and always consider how EM radiation transports energy through space. Start with the total energy striking the detector. Intensity tells you power per unit area, so total power absorbed is $P = I \times A = 800 \times 0.020 = 16 \text{ W}$ . Over 30 seconds, the incident energy is $E_{incident} = 16 \times 30 = 480 \text{ J}$ . Since only 25% converts to electrical energy, $E_{electrical} = 0.25 \times 480 = 120 \text{ J}$ . As for how the energy arrived: electromagnetic radiation carries energy through oscillating electric and magnetic fields — no medium or charged particles required. This makes B the correct answer. A is wrong on two counts: the 480 J figure ignores the 25% efficiency, and the mechanism described — charged particles carrying energy — is not how EM radiation works. EM waves are self-propagating field disturbances, not streams of charged particles. C gets the mechanism partially right (fields carry energy) but the calculation is completely off, producing 20 J from nowhere. It also incorrectly dismisses the magnetic field; both the electric and magnetic fields together transport energy in an EM wave, quantified by the Poynting vector. D invents a nonsensical idea that efficiency multiplies energy — efficiency always reduces usable output, never amplifies it. The 3,000 J figure has no valid calculation behind it. Remember: efficiency is always ≤ 100%, and EM energy is always carried by coupled electric and magnetic fields — not particles.
A radio wave has $E$ along +x and $B$ along +z. Its energy travels toward:
Answer: $-y$ Energy flow follows the Poynting vector, E x B. With E along +x and B along +z, x cross z points along -y, so the wave travels toward -y. The tempting mistake is choosing a field direction like +x or +z, but you must cross E into B, not just look at the fields.
A probe moves upward with $600 J$ of kinetic energy after its engine shuts off. Before reaching a higher altitude, gravity does $-450 J$ of work on the probe. Air resistance is negligible. Which energy description is correct at the higher altitude?
Answer: The probe has $150 J$ of kinetic energy, while the gravitational potential energy of the probe-planet system has increased by $450 J$ . When a moving object experiences a net force, the work-energy theorem tells you exactly how energy transforms. Here, the key is tracking how gravity's work redistributes energy within the probe-planet system — not destroys it. Gravity does $-450 \text{ J}$ of work on the probe. By the work-energy theorem, the probe's kinetic energy changes by that amount: $600 \text{ J} + (-450 \text{ J}) = 150 \text{ J}$ . Simultaneously, whenever gravity does negative work on an object moving upward, that "lost" kinetic energy doesn't disappear — it converts into gravitational potential energy stored in the system. So the system gains exactly $+450 \text{ J}$ of gravitational PE. Answer A correctly captures both facts: $150 \text{ J}$ of KE remaining and $450 \text{ J}$ of PE gained. Answer B is wrong on both counts — adding $450 \text{ J}$ to KE would mean gravity did positive work (which happens when objects fall, not rise), and PE would increase, not decrease. Answer C misunderstands conservation of energy: the gravitational field cannot "remove" energy from the system — it only converts KE to PE. The original $600 \text{ J}$ is never lost, just redistributed. Answer D is contradictory — if gravity adds potential energy, it must be drawing from kinetic energy; you cannot gain PE while KE stays unchanged with no external energy source. A reliable strategy: whenever gravity does work, ask yourself two questions — "How does KE change?" (use the work-energy theorem) and "How does system PE change?" (equal and opposite to work done by gravity). These two changes must account for each other.
A ball of mass $2.0 kg$ is released from rest $5.0 m$ above the ground. Air resistance is negligible, and the gravitational field strength is $10 N/kg$ . Which statement best explains the energy transfer before the ball reaches the ground?
Answer: The ball's gravitational potential energy decreases by $100 J$ , and the gravitational field transfers that energy to the ball as kinetic energy. When a question asks about energy transfer during free fall, your job is to track what happens to each form of energy within the Earth-ball system — not just notice that the ball speeds up. Here's the core idea: gravitational potential energy (GPE) is stored in the Earth-ball system, not in the ball alone. As the ball falls, the gravitational field acts as the agent that converts GPE into kinetic energy (KE). The total mechanical energy stays constant because air resistance is negligible. The GPE lost equals: $\Delta GPE = mgh = (2.0)(10)(5.0) = 100 \text{ J}$ That 100 J doesn't vanish — the gravitational field transfers it directly to the ball as kinetic energy. This makes A the correct answer: GPE decreases by 100 J, and the field transfers that energy to the ball as KE. B is wrong because GPE decreases as the ball falls to a lower height — it doesn't increase just because the force is downward. Force direction and energy direction are different concepts. C is wrong because energy is conserved, not created. The gravitational field doesn't manufacture new energy; it converts existing GPE into KE. The total system energy remains 100 J throughout. D reverses the direction of transfer. When a force causes displacement in the same direction, the field does positive work on the ball, transferring energy to the ball — not away from it. As a study tip: always ask "what form does energy start in, what form does it end in, and what acts as the transfer agent?" This framing will serve you on any energy-conversion question.
A rocket raises a satellite from radius r to radius 2r. The gravitational potential energy:
Answer: Increases Gravitational potential energy is negative, given by -GMm/r. Moving from r to 2r makes the magnitude smaller, so the value goes from -GMm/r to -GMm/(2r), which is a larger number. It increases, becoming less negative, though it never turns positive. The tempting wrong answer is thinking it decreases because the satellite is farther away, but farther means less negative, not smaller.
A charged parallel-plate capacitor is disconnected from its battery and initially stores $0.40 J$ . An external agent slowly doubles the separation between the plates while keeping them parallel. Fringing effects are negligible. Which statement best explains the resulting energy change?
Answer: The capacitance is halved, so the field energy rises to $0.80 J$ ; the external agent transfers $0.40 J$ into the electric field. Whenever a capacitor question involves disconnecting the battery before making a change, your first instinct should be: charge Q is conserved, not voltage. This single insight unlocks the entire problem. When the battery is disconnected, no charge can flow on or off the plates, so $Q$ stays fixed. The capacitance of a parallel-plate capacitor is $C = \varepsilon_0 A/d$ , so doubling the separation $d$ halves the capacitance: $C' = C/2$ . Now use the energy formula written in terms of the conserved quantity: $U = Q^2/(2C)$ . Since $Q$ is unchanged but $C$ is halved, the energy doubles: $U' = Q^2/(2 \cdot C/2) = Q^2/C = 2U = 0.80\text{ J}$ . That extra $0.40\text{ J}$ had to come from somewhere — the external agent pulling the plates apart works against the attractive electrostatic force, injecting energy into the field. So D is correct. A is wrong on two counts: the battery is disconnected (it supplies nothing), and voltage is not fixed when $Q$ is conserved and $C$ changes. B correctly identifies that capacitance halves, but then incorrectly applies the formula $U = \tfrac{1}{2}CV^2$ as if voltage were constant — a trap for students who forget which quantity is conserved. When charge is fixed, energy increases as $C$ decreases. C incorrectly claims charge is halved; charge cannot change because the capacitor is isolated. Study tip: Always ask "what's conserved?" before substituting into an energy formula. Disconnected battery → fixed $Q$ → use $U = Q^2/(2C)$ . Connected battery → fixed $V$ → use $U = \tfrac{1}{2}CV^2$ . These two scenarios give opposite energy trends when $C$ changes.
A proton enters a uniform magnetic field with its velocity perpendicular to the field. It follows a curved path and then enters a uniform electric field directed parallel to its velocity. Which comparison correctly explains the proton's energy changes?
Answer: The magnetic field changes the proton's direction without changing its kinetic energy, while the electric field increases its kinetic energy. When a charged particle moves through fields, the key question is always: does the force do work on the particle? Work — and therefore kinetic energy change — only occurs when a force has a component parallel to the displacement. A magnetic force is always perpendicular to a particle's velocity (given by $\vec{F} = q\vec{v} \times \vec{B}$ ). Because the force is 90° from the motion at every instant, it does zero work: $W = \vec{F} \cdot \vec{d} = 0$ . The magnetic field continuously redirects the proton — producing that curved path — but never speeds it up or slows it down. Speed stays constant; kinetic energy is unchanged. When the proton then enters the electric field directed parallel to its velocity, the electric force acts along the direction of motion, doing positive work and directly increasing the proton's kinetic energy. This makes C the correct answer. A is wrong because it assumes any force on a moving object increases kinetic energy — it ignores the requirement that the force must have a component along the velocity. B has the situation completely reversed: it credits the magnetic field with an energy change and strips the electric field of its energy-transferring role — exactly backward from what actually happens. D goes too far in the wrong direction; electric fields absolutely can and do transfer energy to charged particles through the work-energy theorem. A reliable study tip: whenever you see a magnetic field question, ask yourself "is the force perpendicular to velocity?" If yes, kinetic energy is unchanged — this is almost always a trap on exam questions pairing magnetic and electric fields.
After a switch closes, a distant charge feels a force. What transfers the energy?
Answer: A moving electromagnetic field Closing the switch changes the current, which changes the electromagnetic field around the wire. That changing field travels at light speed and carries energy to the distant charge, so the charge feels a force. The tempting wrong answer is electrons moving in the wire: electrons drift slowly and never leave the wire to reach the distant charge, so they cannot transfer the energy.
A student pulls a bar magnet away from a conducting coil at constant speed. The coil is connected to a resistor, which becomes warmer. The student notices that maintaining the magnet's speed requires a pulling force. Which statement best accounts for the energy transfer?
Answer: The student's work is transferred through the changing electromagnetic field and ultimately becomes thermal energy in the resistor. Whenever you see a question combining electromagnetic induction, a resisting force, and heat generation, you should immediately think about energy conservation — specifically, where energy originates and how it transforms. Here's the chain of events: as you pull the magnet away, the magnetic flux through the coil decreases. By Faraday's law, this induces an EMF, which drives a current through the resistor, converting electrical energy into heat. But that energy has to come from somewhere. By Lenz's law, the induced current creates a magnetic field that opposes the magnet's motion — this is exactly the pulling force the student feels. To maintain constant speed, the student must do work against that opposing force. That mechanical work is what fuels the entire process. C correctly traces this path: the student's work → changing electromagnetic field → electrical energy → thermal energy in the resistor. A reverses causality entirely. Heat from the resistor does not travel back through the field to create the force — energy flows one way, from mechanical input to thermal output, not the reverse. B is a serious misconception: magnetic fields do not supply "unlimited energy." Fields are intermediaries that transfer energy — they don't generate it. The student is the energy source, not the field. D implies the coil creates energy from flux change, which violates conservation of energy. The opposing force doesn't cap some runaway energy production — it's simply the mechanical reaction that channels the student's work into the circuit. Your takeaway: always ask "where does the energy originate?" In induction problems, the mechanical agent (person, motor, etc.) is almost always the primary energy source.

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