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D3.2 · Compare gravitational, electric, and magnetic fields

Learn to compare gravitational, electric, and magnetic fields through clear examples and targeted practice.

Ontario Grade 12 Physics

Gravitational, Electric, and Magnetic Fields

SPH4U study topic D3.2

A field describes how an object can experience a force because of its location or motion. A force is a vector: it has both magnitude and direction. Mass, charge, speed, and field strength are scalars; each has magnitude but no direction. Before analyzing a field, identify the system, the reference frame, and the positive direction. Here, the system is the object experiencing a force, the reference frame is the surrounding ground or laboratory, and a chosen positive direction is used consistently to describe vectors. The three fields in this lesson are gravitational, electric, and magnetic. They can all exert forces without the objects needing to touch, but they differ in what produces them and how they act.

What you will learn

1. Prerequisite bridge: forces and field diagrams

In earlier physics, a force was treated as an interaction that can change an object's motion. A field is a way to describe where that interaction can occur. The field exists in the region around its source; a suitable object placed there experiences a force. The source and the object experiencing the force are not the same role.
A field diagram uses field lines and arrows to represent direction. The arrow at a point gives the direction of the field there. Closer spacing of lines indicates a stronger field in a qualitative diagram, but the lines are not physical paths that an object must follow. A uniform field has the same strength and direction throughout the region shown; parallel, evenly spaced arrows can represent one.
A field is a vector because its direction matters. Field strength is its magnitude. When a force is calculated, its direction must still be stated separately or inferred from the field and the properties and motion of the object.

2. Three fields, three kinds of interaction

A gravitational field is produced by mass. It acts on objects with mass, and the gravitational force is attractive. Near Earth's surface, the gravitational field points downward, toward Earth's centre. Gravitational field strength is force per unit mass. Its SI unit is newtons per kilogram, equivalent to metres per second squared.
An electric field is produced by electric charge. It acts on charged objects. The electric field direction is defined as the direction of force on a small positive test charge. A positive charge experiences force along the field; a negative charge experiences force opposite the field. Electric forces can attract or repel, depending on the signs of the interacting charges. Electric field strength is force per unit charge, measured in newtons per coulomb.
A magnetic field is produced by magnets and by moving electric charges, including electric currents. It acts on magnets and on moving charges or currents. For a moving charged particle, the magnetic force depends on its speed and on the angle between its velocity and the magnetic field. The force is zero if the particle is at rest or moves parallel to the field. When the force is nonzero, it is perpendicular to both the particle's velocity and the magnetic field. Magnetic field strength is measured in teslas.
For a positive moving charge, use the right-hand rule: point your fingers in the direction of the velocity and curl them toward the magnetic-field direction; your thumb indicates the force direction. For a negative charge, reverse that direction. Magnetic field diagrams use arrows or field lines to show the field direction. Around a bar magnet, outside the magnet, the lines point from north to south.
g=Fg/m,E=Fe/q,FB=∣q∣vBsin⁡θg=F_g/m,\quad E=F_e/q,\quad F_B=|q|vB\sin\theta

3. Compare field quantities and use them

The relationships in the previous section define field strength using the force on a suitable object. In these relationships, FgF_g is gravitational force, FeF_e is electric force, FBF_B is magnetic force, mm is mass, qq is charge, vv is speed, and θ\theta is the angle between velocity and magnetic field. The magnitude of charge is used in the magnetic-force equation; direction is found separately.
The first two relationships can be rearranged to find force: gravitational force equals mass times gravitational field strength, and electric force equals charge times electric field. For electric force, a negative charge makes the force direction opposite to the electric field. For the magnetic relationship, the sine factor accounts for the angle: a right angle gives the greatest force for fixed charge, speed, and field, while parallel motion gives zero force.
The table summarizes what each field acts on and how its direction is defined. Use the field's SI unit when reporting a value. Check whether the question asks for the field itself or the force on a particular object.
Fg=mg,Fe=qEF_g=mg,\quad F_e=qE

4. A comparison model

All three fields describe interactions across a region and can be represented with directions. Their key difference is the property that couples an object to the field: mass for gravity, charge for an electric field, and magnetic moment or moving charge/current for magnetic effects. In this course comparison, the magnetic force on a moving charged particle is the useful model.
The force direction is not identical across the three cases. Gravitational force points with the gravitational field for a positive mass. Electric force follows the electric field only for a positive charge. Magnetic force on a moving charge is perpendicular to the field and motion, and its direction also depends on charge sign. These distinctions explain why a field diagram must be combined with information about the object before predicting its force.

Field comparison

FieldSource and what it acts onForce directionField-strength unit
GravitationalMass; acts on massAttractive; force follows field for positive massN/kg
ElectricCharge; acts on chargeAlong field for positive charge; opposite for negativeN/C
MagneticMagnets and moving charges or currents; acts on magnets and moving charges or currentsFor a moving charge, perpendicular to velocity and field; reverse for negative chargeT

Worked example

Gravitational field strength

A 2.40 kg object experiences a downward gravitational force of 23.5 N near a planet's surface. Find the gravitational field strength.
  1. Set the system and direction
    The system is the object. Use the planet's surface as the reference frame and choose downward as positive. The known values are the object's mass and gravitational force; the unknown is field strength.
  2. Apply the definition
    Gravitational field strength is gravitational force divided by mass. The force and mass must be in newtons and kilograms so the result is in newtons per kilogram.
    g=Fgm=23.5 N2.40 kg=9.79 N/kgg=\frac{F_g}{m}=\frac{23.5\ \mathrm{N}}{2.40\ \mathrm{kg}}=9.79\ \mathrm{N/kg}
  3. State the vector result
    The magnitude is three significant figures. Since the force points downward and the mass is positive, the field also points downward.
Answer: The gravitational field strength is 9.79 N/kg9.79\ \mathrm{N/kg} downward.
Check: The units reduce to force per mass. The direction agrees with the measured force on a positive mass, and a field of this scale is reasonable near a planet's surface.

Worked example

Electric force on a negative charge

A charge of −3.0 μC-3.0\ \mu\mathrm{C} is placed in a uniform electric field of 2.5×104 N/C2.5\times10^4\ \mathrm{N/C} directed east. Find the force on the charge.
  1. Set the system and direction
    The system is the charged particle, viewed in the laboratory frame. Take east as positive. The known values are charge and field; the unknown is electric force.
  2. Convert and use the field relationship
    Convert microcoulombs to coulombs. Electric force is charge multiplied by electric field, so the negative charge gives a negative signed result for the east-positive axis.
    q=−3.0×10−6 C,Fe=qE=(−3.0×10−6 C)(2.5×104 N/C)=−7.5×10−2 Nq=-3.0\times10^{-6}\ \mathrm{C},\quad F_e=qE=(-3.0\times10^{-6}\ \mathrm{C})(2.5\times10^4\ \mathrm{N/C})=-7.5\times10^{-2}\ \mathrm{N}
  3. Interpret the sign
    The negative result means the force points opposite east, which is west. Keep two significant figures, matching the given data.
Answer: The force is 7.5×10−2 N7.5\times10^{-2}\ \mathrm{N} west.
Check: Coulombs cancel, leaving newtons. A negative charge experiences force opposite the electric field, so west is consistent with the stated field direction.

Worked example

Magnetic force and orientation

A proton moves at 4.0×105 m/s4.0\times10^5\ \mathrm{m/s} north through a 0.30 T0.30\ \mathrm{T} magnetic field directed east. Find the magnetic-force magnitude and direction. Use the proton charge magnitude 1.60×10−19 C1.60\times10^{-19}\ \mathrm{C}.
  1. Set the system and direction
    The system is the proton in the laboratory frame. Its velocity is north and the field is east, so the angle is 90∘90^\circ. The unknowns are force magnitude and direction.
  2. Calculate the magnitude
    Use the magnetic-force relationship for a moving charge. The sine of a right angle is one, so the force has its maximum magnitude for these values.
    FB=∣q∣vBsin⁡θ=(1.60×10−19 C)(4.0×105 m/s)(0.30 T)sin⁡90∘=1.9×10−14 NF_B=|q|vB\sin\theta=(1.60\times10^{-19}\ \mathrm{C})(4.0\times10^5\ \mathrm{m/s})(0.30\ \mathrm{T})\sin90^\circ=1.9\times10^{-14}\ \mathrm{N}
  3. Find the direction
    For a positive charge, the right-hand rule with velocity north and field east gives a force vertically downward. The proton is positive, so do not reverse the direction.
Answer: The magnetic force is 1.9×10−14 N1.9\times10^{-14}\ \mathrm{N} downward.
Check: Using 1 T=1 N s/(C m)1\ \mathrm{T}=1\ \mathrm{N\,s/(C\,m)}, the units reduce to newtons. The perpendicular velocity and field produce a nonzero force, and the right-hand-rule direction matches the stated orientation.

Common mistakes and how to avoid them

Assuming the electric force always follows the electric-field direction.
Correction: It follows the field for a positive charge and points opposite the field for a negative charge.
Applying the magnetic-force equation to a stationary charged particle.
Correction: A stationary charge has no magnetic force in this model because its speed is zero.
Treating field strength as the same quantity as force.
Correction: Field strength describes force per mass or per charge, while force depends on both the field and the object placed in it.
Giving only a magnitude when a question asks for a force vector.
Correction: Report the magnitude and a direction, using the field direction, charge sign, and motion as needed.

Lesson summary

Check your understanding

Question 1

An electron is placed at rest in an electric field pointing north. Which way is the electric force?
  1. North
  2. South
  3. There is no electric force
  4. correctIndex and explanation
Show answer and explanation
South
An electron has negative charge, so its electric force is opposite the field. It points south.

Question 2

A positive charged particle moves parallel to a magnetic field. What is the magnetic-force magnitude in this model?
  1. Zero
  2. The maximum possible for that speed and field
  3. Equal to its electric force
  4. correctIndex and explanation
Show answer and explanation
Zero
Parallel motion makes the angle zero, and the sine of zero is zero.

Key terms

Field
A description of how a source can exert an interaction throughout a region.
Field strength
The magnitude of a field, expressed in a unit appropriate to the interaction.
Uniform field
A field with constant direction and strength throughout the region considered.
Test charge
A small positive charge used to define the direction of an electric field.
Reference frame
The viewpoint and coordinate system used to describe position, motion, and direction.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation D3.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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