IB Physics HL D.2 Electric and magnetic fields Questions

SLSL HLHL

Practise IB Physics HL D.2 by analysing charge interactions, electric and magnetic fields, potential energy and electric potential, work, and equipotential patterns.

173 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Explain like-charge repulsion and unlike-charge attraction before combining force directions; same-sign ion beams spread through mutual repulsion.
Between equal positive charges, a displaced positive charge feels a restoring force towards the midpoint; at the exact midpoint the forces cancel.
State Coulomb’s law for point charges: force is proportional to the product of charge magnitudes and inversely proportional to separation squared.
Use SI charges and source-centre distances in F = k|q₁q₂|/r²; tripling separation at fixed charges gives one ninth the force.
For a supplied classical circular atomic model, equate electrical attraction to mv²/r and include the actual nuclear charge factor.
At a zero-field point, verify that the source fields oppose, then equate kQ/r² using each full source-to-point distance.
For an observation point beyond two positive sources, add their fields using the separate source distances and give the final charge unit.
Balance signed charge in a fully charge-labelled reaction to infer an unknown, retaining its sign and elementary-charge unit.
Identify Millikan as the scientist who showed that electric charge is quantized; this evidence tests recognition rather than an experimental account.
When a neutral sphere is connected to a positively charged sphere, explain that electrons leave the neutral sphere, leaving it positive.
Define electric field strength as force per unit charge on a small positive test charge; use F = qE for the force on a specified charge.
Use the full ionic charge in F = qE; for an electron reverse the field direction to obtain its force direction.
Calculate point-source field magnitude with E = k|Q|/r² and state direction separately; do not let a negative source make a magnitude negative.
For symmetric source fields, cancel opposing components and add aligned components; an equilateral pair gives 2E cos30° along its symmetry axis.
Between +Q and −Q separated by d, midpoint fields point the same way and add to 8kQ/d²; scalar potential cancellation does not cancel the field.
For point-source field ratios use the distance from the source: if YZ = 2XY on the same ray, XZ = 3XY and EY/EZ = 9.
An electron repelled by a negative sphere gains speed while its acceleration decreases as it moves into a weaker radial field.
Along a constant-radius equipotential in the supplied radial field, field and force magnitudes remain unchanged even though direction can change.
Recognize that field-line density represents field strength: closer line spacing indicates a stronger field.
For uniform plates use the full potential difference divided by the full gap; a midpoint does not halve d in E = |ΔV|/d.
Convert kilovolts and gap units before dividing; a 20 kV potential change across 5.0 cm corresponds to 4.0 × 10⁵ N/C.
Use E = V/d with F = qE to obtain q = Fd/V; for a suspended particle balance the signed electric force against its weight.
Voltage or plate-gap changes can break a charged particle’s equilibrium; applying a magnetic field alone gives no initial force when it is at rest.
When the stem models a conductor field as uniform, first obtain its voltage from V = IR, then divide by the stated length.
Justify a cloud-ground plate model using a flat or parallel cloud base, or a cloud width large compared with its height.
Use the stated breakdown-threshold voltage and plate separation to find the discharge field; do not substitute the supply emf.
Draw electric field arrows from positive to negative; for two positive charges, draw outward curves bending away from the region between them.
Plate-field sketches need straight, evenly spaced normal lines with consistent arrows; follow each scheme’s minimum line count and edge instructions.
For a positive point charge above a conducting plate, field lines leave the charge and meet the plate normally.
Infer an undeflected electron’s electric field by balancing force directions, then reversing the electric force because the charge is negative.
Sketch concentric magnetic field circles around a straight wire: conventional current out of the page gives anticlockwise lines, current in gives clockwise.
Reverse electron flow to obtain conventional current before applying the magnetic right-hand rule.
For multiple wires, combine local magnetic vectors using actual current directions and symmetry; do not add current arrows as field vectors.
For facing north poles, magnetic field lines leave each pole and bend away from the gap; they do not connect north to north.
A rod’s own magnetic field can reinforce the imposed field on one side and oppose it on the other; identify both directions and sides.
In the supplied Earth-field experiment, an east-west coil axis makes its field perpendicular to the north-south Earth field, enabling a deflection measurement.
In the stated helium-ion model, Ep = −2ke²/r and Ek = ke²/r, so the total energy is −ke²/r.
For that supplied classical orbit, radiation loss makes total energy more negative and therefore reduces its radius.
Define electric potential using work per unit positive test charge from infinity, and add source potentials as signed scalars.
Evaluate potential and field separately: symmetric sources can have zero field with either zero or nonzero scalar potential.
Apply V = kQ/r outside a sphere with centre distance; use constant Vr for the same source and Q = VR/k at its surface.
For conducting-sphere potential graphs, show a constant nonzero interior and a continuous inverse-radius exterior approaching zero.
Between equal positive point charges, sketch a nonzero positive bowl with charge asymptotes; finite conductors instead have flat interior potentials.
For connected, well-separated spheres, combine equal final kq/R potentials with conserved total charge to find the charge sharing.
For square-corner potentials use the centre-to-corner distance, distinguishing a given side from a diagonal before summing source contributions.
A two-charge potential minimum has zero gradient; infer a charge ratio by equating the opposing inverse-square fields at that point.
Use E = −dV/dr: take a local tangent for field at a point, and interpret field-graph area as a potential drop with the correct sign.
In a uniform field, potential change depends only on the field-parallel displacement and increases opposite to the field.
Use ΔU = qΔV and Wfield = −qΔV; retain the charge sign and distinguish work by the field from external work.
Electrostatic work depends on endpoint potentials; equipotential path segments add no energy change and a closed return path has zero net work.
Convert electrical energy to kinetic energy using the full particle charge and the crossed voltage fraction; a plate midpoint gives half the full voltage.
Test reachability or stopping by comparing the complete qΔV barrier with available kinetic energy, then give an explicit conclusion.
For an electron, U = −eV is greatest at the lowest potential; do not rank its potential energy as if it were positively charged.
For a supplied relativistic energy graph, calculate current-syllabus γ, read the given kinetic energy and divide by charge to obtain the voltage.
In the proton-nucleus approach problem, explain repulsion, the need to reach the short nuclear-force range, and the initial kinetic energy needed.
Identify uniform plate equipotentials as parallel planes; position specified potential levels in proportion to their fraction of the voltage gap.
A symmetric closed two-source contour pattern can represent two equal masses or equal same-sign charges; use the entire native pattern.
Find a field normal towards lower potential, then reverse it for a negative charge’s acceleration; retain any pre-existing velocity in the trajectory.
A given radial trajectory crosses a sphere’s equipotentials normally; this does not imply every moving charge has velocity along the electric field.

Covered syllabus points

D.2.1—Electric charge forces D.2.2—Coulomb’s law D.2.3—Charge conservation D.2.4—Millikan experiment D.2.5—Charge transfer D.2.6—Electric field strength D.2.7—Electric field lines D.2.8—Field-line density D.2.9—Parallel-plate field D.2.10—Magnetic field lines D.2.11 (HL)—Electric potential energy D.2.12 (HL)—Two-charge potential energy D.2.13 (HL)—Electric potential as scalar D.2.14 (HL)—Electric potential D.2.15 (HL)—Electric potential gradient D.2.16 (HL)—Work in electric fields D.2.17 (HL)—Electric equipotentials D.2.18 (HL)—Equipotentials and electric fields

Exam points

Explain spreading of a beam of same-sign ions through mutual electrostatic repulsion; state both their common charge sign and the repulsive force.
Use repulsion between like charges and attraction between unlike charges before combining the force directions on a test charge.
For symmetric charge arrangements, cancel opposing components and add those pointing the same way using the actual signs and positions.
Between two equal positive charges, a displaced positive charge feels a resultant towards the midpoint; at the midpoint the two forces cancel.

[Maximum number: 2]

Ion-thrust engines can power spacecraft. In this type of engine, ions are created in a chamber and expelled from the spacecraft. The spacecraft is in outer space when the propulsion system is turned on. The spacecraft starts from rest.

The mass of ions ejected each second is $6.6 \times 10^{-6} \mathrm{~kg}$ and the speed of each ion is $5.2 \times 10^{4} \mathrm{~m} \mathrm{~s}^{-1}$ . The initial total mass of the spacecraft and its fuel is 740 kg . Assume that the ions travel away from the spacecraft parallel to its direction of motion.

In practice, the ions leave the spacecraft at a range of angles as shown.

Outline why the ions are likely to spread out.

Exam points

State that the force between two point charges is proportional to their charge product and inversely proportional to separation squared.
Use F = k|q₁q₂|/r² with charge in coulombs and centre-to-centre separation in metres; retain the squared distance.
For fixed charges, increasing separation by a factor of three reduces Coulomb force to one ninth.
In a stated classical circular orbit, equate Coulomb attraction to mv²/r and include the correct nuclear charge factor.
At a zero-field point, first check the two fields oppose, then equate kQ/r² using each source-to-point distance.
Find unknown charge ratios from squared distances; use the full geometric distance and round only the final result as requested.
For positive sources with an observation point beyond both, add the fields and use each sphere centre to locate the point.
Keep charge sign consistent with field cancellation: a zero field between two charges requires the charges to have the same sign.

[Maximum number: 2]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

The diagram shows two isolated electrons, X and Y , initially at rest in a vacuum. The initial separation of the electrons is 5.0 mm . The electrons subsequently move apart in the directions shown.

Show that the initial electric force acting on each electron due to the other electron is approximately $9 \times 10^{-24} \mathrm{~N}$ .

(c) (i) use of $F=\frac{\left(1.60 \times 10^{-19}\right)^{2}}{4 \pi \varepsilon_{0}\left(5.0 \times 10^{-3}\right)^{2}}$ or $F=\frac{\left(1.60 \times 10^{-19}\right)^{2}}{\left(5.0 \times 10^{-3}\right)^{2}} \times 8.99 \times 10^{9}$ ;
$9.2 \times 10^{-24} \mathrm{~N} ;$

Exam points

Sum signed charges before and after a fully labelled reaction to find an unknown charge; retain its sign and elementary-charge unit.

[Maximum number: 1]

Deuterium, ${ }_{1}^{2} \mathrm{H}$ , undergoes fusion according to the following reaction.

{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{1}^{3} \mathrm{H}+\mathrm{X}

Particle Y is produced in the collision of a proton with a $\mathrm{K}^{-}$ in the following reaction.

\mathrm{K}^{-}+\mathrm{p}^{+} \rightarrow \mathrm{K}^{0}+\mathrm{K}^{+}+\mathrm{Y}

The quark content of some of the particles involved are

\mathrm{K}^{-}=\overline{\mathrm{u}} \mathrm{~s} \quad \mathrm{~K}^{0}=\mathrm{d} \overline{\mathrm{~s}}

Identify, for particle Y , the

charge.

Exam points

Identify Millikan as the scientist who first showed that electric charge is quantized.

[Maximum number: 1]

Who was the first scientist to show that electric charge is quantized?

Coulomb

Millikan

Planck

Rutherford

Exam points

When an initially neutral sphere is connected to a positively charged sphere, describe electrons leaving the neutral sphere so it becomes positive.

[Maximum number: 1]

A conducting sphere has radius 48 cm . The electric potential on the surface of the sphere is $3.4 \times 10^{5} \mathrm{~V}$ .

The sphere is connected by a long conducting wire to a second conducting sphere of radius 24 cm . The second sphere is initially uncharged.

Describe, in terms of electron flow, how the smaller sphere becomes charged.

Exam points

Define electric field strength as force per unit charge on a small positive test charge, including the positive-charge convention.
Use F = qE for electric force; include the full ionic charge, such as +2e, rather than always substituting one elementary charge.
A positive charge feels force along the electric field; an electron feels force opposite to it.
Use E = k|Q|/r² for point-charge field magnitude and state its direction separately from the sign used in a magnitude calculation.
Outside a conducting sphere, use distance from its centre; distinguish the sphere radius used for surface potential from the field-point radius.
For opposite unequal charges, a zero-field point lies outside the pair on the weaker-charge side, where the two fields oppose.
At the midpoint between +Q and −Q separated by d, add both fields at distance d/2 to obtain a magnitude of 8kQ/d².
Resolve equal source fields into components: cancel those opposed by symmetry and add those in the resultant direction.
For an equilateral two-charge arrangement, calculate one field first, then use 2E cos30° along the symmetry axis.
Compare vector contributions before ranking field strength; between opposite charges, their fields point the same way and add.
For a released electron repelled by a negative sphere, speed increases while acceleration decreases as the field weakens with distance.
With negligible air resistance, constant-velocity charged droplets have balanced electric force and weight, giving qE = mg.
Use the distance from the source in inverse-square ratios: if XZ = 3XY, the field at Y is nine times the field at Z.

[Maximum number: 4]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

(a)

Define electric field strength.

[ 2 ]

(b)

(b) The diagram shows a pair of horizontal metal plates. Electrons can be deflected vertically using an electric field between the plates.

[ 2 ]

(i)

Calculate the force on an electron between the plates when the electric field strength has a value of $2.5 \times 10^{3} \mathrm{NC}^{-1}$ .

[ 2 ]

$F=2.5 \times 10^{3} \times 1.6 \times 10^{-19}$ ;
$4.0 \times 10^{-16} \mathrm{~N}$ ;

Marking guidance:

Award [2] for a bald correct answer.

Exam points

Draw electric field arrows from positive towards negative charge, and infer a source sign from how its field lines connect.
For two positive charges, field lines point outwards and bend away from the region between them; do not join one positive charge to the other.
For an equal dipole on a line, compare the opposing fields outside the pair and add them between the charges to determine direction.
Between parallel plates, draw straight, evenly spaced lines normal to both plates, with arrows from the positive plate to the negative plate.
Include curved edge fields when the scheme requires them; for sketches that explicitly ignore edge effects, follow that instruction.
At a conducting plate, electric field lines meet the surface at right angles; retain the outward direction from a nearby positive charge.
For an undeflected electron in crossed fields, balance the force directions and reverse the electric force direction to obtain the field arrow.
When a plate field accelerates an electron upwards, the field points downwards; draw downward arrows consistently and no upward arrows.
Combine the field vectors from each signed corner charge before selecting a resultant arrow; a diagonal source can cancel one component.

[Maximum number: 2]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

(b) The diagram shows a pair of horizontal metal plates. Electrons can be deflected vertically using an electric field between the plates.

Draw the shape and direction of the electric field between the plates on the diagram.

Exam points

Recognize that electric field lines represent field strength through their density: closer spacing indicates a stronger field.

[Maximum number: 1]

Electric field lines

can cross each other.

are parallel to equipotential surfaces.

are directed from negative to positive charges.

show field strength by their density.

Exam points

Use E = |ΔV|/d with the total potential difference and full plate separation; a midpoint does not halve the separation in this formula.
For plates at −150 V and +150 V separated by 2.0 m, use a 300 V difference and direct the 150 V/m field towards the negative plate.
Convert kilovolts and centimetres or millimetres before division; a 20 kV change across 5.0 cm gives 4.0 × 10⁵ N/C.
In the uniform region between plates, field strength is constant while potential changes linearly with perpendicular distance.
Combine F = qE with E = V/d to obtain q = Fd/V; for equilibrium with weight, use q = mgd/V.
Use the charge sign when inferring field direction from equilibrium: a suspended negative particle needs an upward force and a downward field.
Changing plate voltage or separation changes the electric force and can break equilibrium; a magnetic field exerts no force on a charge initially at rest.
For a cloud-ground parallel-plate model, state a flat or parallel cloud base, or a cloud width large compared with the separation.
Use the specified discharge-threshold voltage in E = V/d; distinguish it from the power supply emf or a charging-time graph.

[Maximum number: 1]

Two oppositely charged parallel plates are a distance 8.0 cm apart. The potential difference between the plates is 120 V . An alpha particle is placed on the positively charged plate and released from rest. Gravity is ignored.

Calculate the electric field between the plates.

$E=<\frac{120}{8.0 \times 10^{-2}}=>1.5 \times 10^{3} N C^{-1} \boldsymbol{O} \boldsymbol{R} \mathrm{Vm}^{-1}$

Exam points

Around a straight wire, sketch concentric circles centred on the wire, with arrows determined by conventional current.
Use the right-hand rule: current out of the page gives anticlockwise field lines, while current into the page gives clockwise lines.
Meet each drawing scheme’s line-count and spacing requirements; for the outward-current sketch use three circles with increasing outward gaps.
Reverse electron-flow direction to obtain conventional current before using the magnetic-field right-hand rule.
For equal wire currents, combine the local field vectors; at a midpoint equal opposing contributions cancel.
Infer each wire’s current from the actual resultant direction and geometry; equal distances allow equal field components to cancel.
For two facing north poles, field lines leave each pole and bend away from the space between them; do not connect north to north.
Explain unequal fields on a rod’s two sides by naming the rod’s own field, its direction on each side, and whether it adds to the imposed field.

[Maximum number: 2]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

The diagram shows two isolated electrons, X and Y , initially at rest in a vacuum. The initial separation of the electrons is 5.0 mm . The electrons subsequently move apart in the directions shown.

The diagram shows Y as seen from X , at one instant. Y is moving into the plane of the paper. For this instant, draw on the diagram the shape and direction of the magnetic field produced by Y .

[Maximum number: 1]

An electron is fixed in position in a uniform electric field. What is the position for which the electrical potential energy of the electron is greatest?

Exam points

Use Ep = kq₁q₂/r with signed charges: an electron and a +2e helium nucleus have potential energy −2ke²/r.
In the stated classical helium-ion orbit, use the given speed relation to find Ek = ke²/r, then add Ep = −2ke²/r to obtain Etot = −ke²/r.
When the stated classical orbit loses energy by radiation, total energy becomes more negative; Etot = −ke²/r then implies a smaller radius.

[Maximum number: 4]

(a)

In a classical model of the singly-ionized helium atom, a single electron orbits the nucleus in a circular orbit of radius r.

[ 4 ]

(i)

Hence, deduce that the total energy of the electron is given by $E_{\mathrm{TOT}}=-\frac{k e^{2}}{r}$ .

[ 2 ]

uses (a)(i) to state $E_{\mathrm{k}}=\frac{k e^{2}}{r} \mathcal{OR}$ states $E_{\mathrm{p}}=-\frac{2 k e^{2}}{r}$ adds $« E_{\mathrm{TOT}}=E_{\mathrm{k}}+E_{\mathrm{p}}=\frac{k e^{2}}{r}-\frac{2 k e^{2}}{r}$ » to get the result

(ii)

In this model the electron loses energy by emitting electromagnetic waves.

Describe the predicted effect of this emission on the orbital radius of the electron.

[ 2 ]

the total energy decreases OR by reference to $E_{\text {TOT }}=-\frac{k e^{2}}{r}$ the radius must also decrease

Award [0] for an answer concluding that radius increases

Exam points

Define electric potential at a point as the work done per unit positive test charge to bring it from infinity to that point.
Add electric potentials as signed scalars: two positive source charges give a positive total potential even where their fields cancel.
Do not infer zero potential from zero field; justify the potential using its scalar sum or the work needed to bring a test charge from infinity.
For equal-distance corner charges, check the signed potential sum and vector-field cancellation separately before selecting both results.

[Maximum number: 2]

A vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.

The centre of the ball, still carrying a charge of $1.2 \times 10^{-6} \mathrm{C}$ , is now placed 0.40 m from a point charge Q . The charge on the ball acts as a point charge at the centre of the ball.
P is the point on the line joining the charges where the electric field strength is zero. The distance PQ is 0.22 m .

Outline, without calculation, whether or not the electric potential at P is zero.

Exam points

Define potential as work per unit positive test charge brought from infinity to the point, with the zero-potential reference explicit.
Use V = kQ/r outside a point charge or spherical conductor, measuring r from the source centre and retaining the charge sign.
For the same isolated sphere, Vr is constant outside: use centre-based distances to infer its radius from surface and external potentials.
At a midpoint between equal positive charges, add both scalar potentials; each source is half the full source separation away.
Sketch potential between equal positive point charges as a symmetric positive bowl, with a nonzero minimum and asymptotes at the charges.
For a charged conducting sphere, potential is constant throughout the interior and continuous at the surface, then falls as 1/r outside.
For two identical charged conducting spheres, show constant interior potentials and a symmetric positive minimum between them.
Connected, well-separated spheres reach equal potentials: combine q₁/R₁ = q₂/R₂ with conservation of total charge.
Add all source potentials at a square centre using the centre-to-corner distance; convert the given side or diagonal before substitution.
At a two-charge potential minimum, the field is zero; equate inverse-square source fields using each distance to infer the charge ratio.
For a positive sphere, explain the outward potential decrease by the smaller work per unit charge needed to reach a more distant point from infinity.
Use the difference of source potentials, kQ(1/rA − 1/rB), to infer an unknown source charge from a known potential difference.

[Maximum number: 2]

A metal sphere is charged positively and placed far away from other charged objects. The electric potential at a point on the surface of the sphere is 53.9 kV .

The electric potential at a point a distance 2.8 m from the centre of the sphere is 7.71 kV . Determine the radius of the sphere.

use of V r= constant $0.40 \mathrm{~m}$

Marking guidance:

Allow [1] max if r+2.8 used to get 0.47 m .

Allow [2] marks if they calculate Q at one potential and use it to get the distance at the other potential.

Exam points

Obtain electric field from the negative tangent gradient of a potential-distance graph; do not use the area under the potential graph.
A constant potential inside a hollow conducting sphere gives zero field there; its external 1/r potential gives a field falling as 1/r².
The magnitude of the potential gradient equals electric field magnitude, so 950 N/C corresponds to a gradient magnitude of 950 V/m.
In a uniform field, use only the displacement component parallel to the field when calculating a potential difference.
Potential increases opposite to the field: if S is Δy above R in a downward field and VR = 0, then VS = +EΔy.
For displacement s along a uniform field, the potential drop has magnitude Es and is independent of the moving particle’s charge.
Area under an E–r graph gives a potential drop; for outward motion in a positive radial field, ΔV is the negative of that area.

[Maximum number: 1]

This question is about electric potential and electric field.

A metal sphere of radius 0.080 m is charged to a potential of 300 V .

State the value of the magnitude of the gradient of electric potential at a distance of 0.16 m from the centre of the sphere.

A6. This question is about the motion of a copper rod in a magnetic field.

A copper rod XY is able to move freely along two horizontal, parallel conducting rails A and B . The conducting rails are in a region of uniform magnetic field that is in a direction perpendicular to the plane of the rails. The rails are connected together at one end by a conducting wire.

In the situation shown, the rod is moved along the rails at a constant speed v by a constant horizontal force of magnitude F.

$950 \mathrm{~V} \mathrm{~m}^{-1}$ ;

Marking guidance:

Allow ECF from (b)(ii).

Exam points

Use ΔU = q(Vfinal − Vinitial); for work done by the electric field, use the opposite sign, q(Vinitial − Vfinal).
Use the full ionic charge and retain the sign of the energy change: a +2e ion moving to lower potential loses electric potential energy.
Convert electric energy to kinetic energy through ΔK = −qΔV when no other force does work.
An electron retarded through 1.2 V loses 1.2 eV of kinetic energy; subtract this before converting the remaining energy into speed.
To stop every photoelectron, arrange a retarding potential-energy barrier at least as large as the maximum emitted kinetic energy.
Compare required qΔV with initial kinetic energy to decide reachability, and state a conclusion consistent with the numerical comparison.
For work from infinity, add the source potentials at the destination and multiply by the charge being brought in.
Compare electrostatic work using endpoint potentials, including any equipotential arcs; path length alone does not set the work.
The total electrostatic work over a closed return path is zero, even when the work on each separate leg is nonzero.
For fixed charge, kinetic-energy gain depends on the crossed potential difference; doubling the gap does not offset halving the voltage.
From a parallel-plate midpoint, use half the full potential difference when calculating the energy gained before reaching a plate.
Use charge in elementary-charge units when converting eV to voltage; a +2e alpha crossing 120 V gains 240 eV.
If a speed is relativistic but an energy-versus-γ graph is supplied, calculate γ, read the energy, then use the charge to find voltage.
For a supplied final total energy expressed as a multiple of rest energy, subtract the rest contribution before equating kinetic gain to |qV|.
For a negative charge, the greatest external work over equal path lengths is along the field, where motion most strongly opposes its electric force.

[Maximum number: 2]

This question is in two parts

Ionized hydrogen atoms are accelerated from rest in the vacuum between two vertical parallel conducting plates. The potential difference between the plates is V. As a result of the acceleration each ion gains an energy of $1.9 \times 10^{-18} \mathrm{~J}$ .

Calculate the value of V.

$\quad V=\frac{1.9 \times 10^{-18}}{1.6 \times 10^{-19}}$ ;
$=12 \mathrm{~V}$ ;

Exam points

In the uniform region between parallel plates, identify equipotential surfaces as planes parallel to the plates and normal to the field.
For 0 V and −1.2 V plates, place the −0.4 V and −0.8 V equipotentials at one third and two thirds of the gap and label them in order.
A symmetric equipotential pattern with closed contours around both sources can represent two equal masses or two equal charges of the same sign.

[Maximum number: 2]

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to the ground state. Photons are emitted and are incident on a photoelectric surface as shown.

The electric potential of the photoelectric surface is 0 V . The variable voltage is adjusted so that the collecting plate is at -1.2 V .

photoelectric surface

On the diagram, draw and label the equipotential lines at -0.4 V and -0.8 V .

Exam points

Draw the electric field normal to the local equipotential and towards lower potential; use the tangent at the actual observation point.
An electron accelerates opposite to the field, towards increasing potential; distinguish its acceleration from an existing velocity.
For a falling negative particle, combine its initial vertical velocity with the horizontal electric acceleration inferred from the potential labels.
A particle explicitly moving radially towards a charged sphere crosses its spherical equipotential surfaces at 90°.
Equal potential steps across equal spacing imply a uniform field and constant electric acceleration; use endpoint potentials for the work.
For a positive charge released from rest, the initial acceleration and motion just after release are normal to the equipotential towards lower potential.

[Maximum number: 1]

A metal sphere is charged positively and placed far away from other charged objects. The electric potential at a point on the surface of the sphere is 53.9 kV .

A small positively charged object moves towards the centre of the metal sphere. When the object is 2.8 m from the centre of the sphere, its speed is $3.1 \mathrm{~ms}^{-1}$ . The mass of the object is 0.14 g and its charge is $2.4 \times 10^{-8} \mathrm{C}$ .

Comment on the angle at which the object meets equipotential surfaces around the sphere.

$90^{\circ}$ / perpendicular

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