D2.3 Solve electric-force, field, energy, and potential problems

Ontario Grade 12 Physics

Gravitational, Electric, and Magnetic Fields

Solving Grade 12 problems with charge, distance, direction, and energy

In SPH3U, you used force diagrams and energy relationships to describe how objects interact and move. This lesson applies those ideas to electric charges. A charge is a property of matter that can be positive or negative. The electric force between charges can attract or repel. An electric field describes the force that a positive test charge would experience at a location. Electric potential energy and electric potential describe energy associated with charge and position. Unless a problem says otherwise, treat the charges as point charges: their sizes are small compared with the distances between them. First choose a system, reference frame, and positive direction. Then identify whether the question asks for a vector, which has magnitude and direction, or a scalar, which has magnitude only.

What you will learn

Use Coulomb’s law to calculate the electric force between point charges and state its direction.
Calculate electric field strength and use the field to determine the force on a charge.
Relate electric potential energy to electric potential and charge.
Check signs, directions, SI units, and the reasonableness of answers.

1. Set up the system and model

A useful system is the set of charges named in the problem. Use a stationary reference frame attached to the page or room, unless motion is part of the question. For a one-dimensional problem, choose a positive direction, such as right. Keep that choice throughout. A positive answer points right; a negative answer points left. In two dimensions, use perpendicular

x

and

y

directions.

Charge is measured in coulombs, symbol

\mathrm{C}

. The elementary charge has magnitude

e=1.60\times10^{-19}\,\mathrm{C}

. An electron has charge

-e

and a proton has charge

+e

. The electric constant is

k=8.99\times10^9\,\mathrm{N\,m^2/C^2}

. Use values supplied by your course when they differ slightly.

Coulomb’s law gives the electric force between two point charges. The force magnitude increases when either charge magnitude increases, and decreases as the square of the separation increases. Like signs repel; opposite signs attract. Force is a vector, so use the signs of the charges to determine the direction after finding the magnitude.

For more than two source charges, find each force on the chosen charge separately, then add the force vectors. In one dimension, assign each force a sign based on its direction. In two dimensions, add horizontal components together and vertical components together.

F=k\frac{|q_1q_2|}{r^2}

State the system, stationary reference frame, positive direction, known values, and unknown.
Convert charge to coulombs and distance to metres before substituting.
Attraction or repulsion determines direction; magnitude alone does not.

2. Electric field and force

Electric field strength,

E

, describes electric force per unit positive charge at a location. It is a vector. Its SI unit is newtons per coulomb,

\mathrm{N/C}

. The field points in the direction a positive test charge would be pushed. A negative charge feels a force opposite the field direction.

The field around a point source charge has a radial direction. It points away from a positive source charge and toward a negative source charge. Its magnitude depends on the source charge and the distance from it. To draw a simple field diagram, mark the source charge and draw a radial arrow at the location of interest, away from a positive source or toward a negative source.

The force on a charge placed in a known field follows from the definition of field strength. The force is a vector. If the charge is positive, force and field have the same direction; if it is negative, they point in opposite directions. Do not confuse the source charge, which creates the field, with the charge placed in that field.

If several source charges create a field at one point, calculate each field contribution and add them as vectors. A field is not the same as force: the field belongs to a location, while the force depends on the charge placed there.

E=\frac{F}{|q|}=k\frac{|Q|}{r^2}

Field direction is defined using a positive test charge.
A negative charge experiences force opposite to the electric field.
The field is a vector; electric potential, introduced next, is a scalar.

3. Electric potential energy and potential

Electric potential energy,

U

, is energy associated with the positions of interacting charges. It is a scalar measured in joules,

\mathrm{J}

. For two point charges separated by distance

r

, the sign of

U

depends on the signs of the charges: like charges have positive potential energy, and unlike charges have negative potential energy when zero energy is defined at very large separation.

Electric potential,

V

, is electric potential energy per unit charge. It is a scalar measured in volts, where

1\,\mathrm{V}=1\,\mathrm{J/C}

. The potential created by a point source charge is positive for a positive source and negative for a negative source. When several source charges contribute, add their potentials using signs. Unlike fields, potentials add as ordinary signed numbers because they are scalars.

The relationship between potential energy and potential is

U=qV

for a charge

q

at a location, with the chosen zero reference understood. Between two locations, the change in potential energy is

\Delta U=q\Delta V

. A positive charge moving to a higher potential gains potential energy. A negative charge at that same potential change loses potential energy.

When a problem asks about work, read carefully which agent does the work. For a change in electric potential energy, the energy change is

\Delta U=U_f-U_i

. If no other energy transfers are involved, work done by the electric force is the negative of this change. Keep the focus on the requested quantity and its sign.

U=k\frac{qQ}{r},\qquad V=\frac{U}{q}=k\frac{Q}{r}

Potential energy and potential are scalars, so they do not have directions.
Use signed charges in energy and potential relationships.
A change means final value minus initial value.

4. A reliable solution and checking method

Begin by listing what is known and what must be found. Sketch the charges and label the separation. For force or field, add arrows showing the chosen positive direction and the expected direction before calculating. This prediction helps catch sign errors.

Write the governing relationship before substitution. Convert all distances to metres and charges to coulombs. Substitute values with units, then round the final result to a sensible number of significant figures based on the given data. Keep extra digits during intermediate steps.

For force and field, report both magnitude and direction, or use a signed component that matches the chosen coordinate direction. For energy and potential, report the signed scalar value when the reference is specified. Check units: force in newtons, field in newtons per coulomb, energy in joules, and potential in volts.

Check reasonableness. Doubling separation should reduce a point-charge force or field magnitude to one quarter. Doubling separation should halve the magnitude of potential or potential energy. Also compare your direction with attraction or repulsion and your sign with the source and test-charge signs.

\Delta U=q\Delta V

Predict direction before calculating a vector quantity.
Use metres and coulombs in the point-charge equations.
Check the final unit, sign, direction, significant figures, and distance dependence.

Worked example

Force between two charges

Two charges,

q_1=+3.0\,\mu\mathrm{C}

and

q_2=-2.0\,\mu\mathrm{C}

, are

0.40\,\mathrm{m}

apart on a horizontal line. Find the force on

q_2

. Take right as positive, with

q_1

to the left of

q_2

.

Define the system
The system is the two stationary point charges in the page’s reference frame. Right is positive. The unknown is the force on $q_2$ . Since the charges have opposite signs, they attract, so the force on $q_2$ points left.
Convert and choose the relationship
Convert microcoulombs to coulombs. Use Coulomb’s law for the force magnitude, then apply the direction found from attraction.
$q_1=3.0\times10^{-6}\,\mathrm{C},\quad q_2=-2.0\times10^{-6}\,\mathrm{C}$
Calculate the magnitude
Substitute the charge magnitudes, separation, and electric constant. The result is a force magnitude; attach the leftward direction afterward.
$F=(8.99\times10^9\,\mathrm{N\,m^2/C^2})\frac{(3.0\times10^{-6}\,\mathrm{C})(2.0\times10^{-6}\,\mathrm{C})}{(0.40\,\mathrm{m})^2}=0.337\,\mathrm{N}$

Answer: To two significant figures, the force on

q_2

is

0.34\,\mathrm{N}

left, or

-0.34\,\mathrm{N}

in the chosen coordinate direction.

Check: The units reduce to newtons. Opposite charges attract, so the leftward result agrees with the diagram. A separation of less than half a metre and microcoulomb charges make a force of a few tenths of a newton reasonable.

Worked example

Field and force on a negative charge

A source charge

Q=-4.0\,\mu\mathrm{C}

is

0.30\,\mathrm{m}

to the left of a point P. Find the electric field at P and the force on a charge

q=-2.0\,\mathrm{nC}

placed there. Take right as positive.

Set the directions
The source charge and point P define the field location in a stationary reference frame. Right is positive. Because the source is negative, the field at P points toward it, which is left.
Calculate the field
Use the point-source field equation for the magnitude. The field points left, so its signed value is negative.
$E=-k\frac{|Q|}{r^2}=-(8.99\times10^9\,\mathrm{N\,m^2/C^2})\frac{4.0\times10^{-6}\,\mathrm{C}}{(0.30\,\mathrm{m})^2}=-4.0\times10^5\,\mathrm{N/C}$
Find the force
Convert the placed charge to coulombs and use $F=qE$ . The negative charge experiences force opposite the field, so the force points right.
$F=(-2.0\times10^{-9}\,\mathrm{C})(-4.0\times10^5\,\mathrm{N/C})=+8.0\times10^{-4}\,\mathrm{N}$

Answer: The electric field is

4.0\times10^5\,\mathrm{N/C}

left. The force on the negative charge is

8.0\times10^{-4}\,\mathrm{N}

right.

Check: The field unit is newtons per coulomb, and charge times field gives newtons. The negative placed charge feels force opposite the leftward field. Both results have the expected directions.

Worked example

Potential and potential energy

A source charge

Q=+5.0\,\mu\mathrm{C}

is at the origin. A charge

q=-2.0\,\mathrm{nC}

is

0.20\,\mathrm{m}

away. Take electric potential to be zero at very large separation. Find the potential at the charge’s location and the pair’s electric potential energy.

Identify the quantities
The stationary system is the source and the placed charge. Potential at the location depends on the source charge, while pair energy depends on both charges. Both quantities are scalars.
Calculate potential
Use the signed source charge in the point-charge potential equation. A positive source gives positive potential.
$V=k\frac{Q}{r}=(8.99\times10^9\,\mathrm{N\,m^2/C^2})\frac{5.0\times10^{-6}\,\mathrm{C}}{0.20\,\mathrm{m}}=2.2\times10^5\,\mathrm{V}$
Calculate potential energy
Use both signed charges in the energy equation. Since the charges have opposite signs, the energy is negative for the stated zero reference.
$U=qV=(-2.0\times10^{-9}\,\mathrm{C})(2.2\times10^5\,\mathrm{V})=-4.5\times10^{-4}\,\mathrm{J}$

Answer: The potential is

+2.2\times10^5\,\mathrm{V}

, and the electric potential energy is

-4.5\times10^{-4}\,\mathrm{J}

.

Check: A coulomb times a volt is a joule. The positive source makes positive potential, while the opposite-sign pair has negative energy with zero at infinite separation. The energy also agrees with

kqQ/r

.

Common mistakes and how to avoid them

Giving a force magnitude but no direction.

Correction: State attraction or repulsion and connect it to the chosen positive direction. A vector answer needs direction.

Using the placed charge’s sign to decide the electric field direction.

Correction: The field direction is set by the source charge and is defined as the direction of force on a positive test charge.

Treating electric potential as a vector or adding it like a force.

Correction: Potential is a signed scalar. Add source potentials as signed numbers; add force and field contributions as vectors.

Using positive magnitudes for every charge in a potential-energy calculation.

Correction: Keep the charge signs in

U=kqQ/r

. The signs determine whether the result is positive or negative.

Leaving distance in centimetres or charge in microcoulombs during substitution.

Correction: Convert distance to metres and charge to coulombs before using the stated SI constant.

Lesson summary

Coulomb’s law calculates the force between point charges; determine direction from attraction or repulsion.
Electric field is force per unit positive charge and points away from positive sources and toward negative sources.
Electric potential energy depends on both charges; electric potential depends on the source and location.
Force and field are vectors. Potential energy and potential are scalars.
Check units, signs, direction, significant figures, and how the result changes with distance.

Check your understanding

Question 1

A positive source charge creates a field at a point to its right. Which way does the field point?

Right, away from the positive source
Left, toward the positive source
The field has no direction because it is a scalar
The direction cannot be known without a test charge

Show answer and explanation

Right, away from the positive source

The electric field points in the direction a positive test charge would be pushed. A positive source repels it, so the field points away from the source.

Question 2

Two point charges are moved from separation

r

to separation

2r

. How does the magnitude of their electric force change?

It becomes twice as large
It becomes half as large
It becomes one quarter as large
It remains unchanged

Show answer and explanation

It becomes one quarter as large

Coulomb’s law has the square of separation in the denominator. Doubling the separation makes the force magnitude one quarter as large.

Question 3

A charge

q=-3.0\,\mathrm{C}

moves through a potential change of

+4.0\,\mathrm{V}

. What is its change in electric potential energy?

$+12\,\mathrm{J}$
$-12\,\mathrm{J}$
$+1.3\,\mathrm{J}$
$-1.3\,\mathrm{J}$

Show answer and explanation

-12\,\mathrm{J}

Use

\Delta U=q\Delta V

. The product is

(-3.0\,\mathrm{C})(+4.0\,\mathrm{V})=-12\,\mathrm{J}

. The negative charge’s potential energy decreases.

Key terms

Point charge: A model that treats a charged object as if all its charge were concentrated at one location.
Electric force: The push or pull between electric charges. It is a vector measured in newtons.
Electric field: Electric force per unit positive charge at a location. It is a vector measured in newtons per coulomb.
Electric potential energy: Energy associated with the positions of interacting charges, measured in joules.
Electric potential: Electric potential energy per unit charge at a location, measured in volts.
Scalar: A quantity with magnitude only, such as energy or potential.
Vector: A quantity with both magnitude and direction, such as force or electric field.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation D2.3. 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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D2.3 · Solve electric-force, field, energy, and potential problems

Ontario Grade 12 Physics

Solving Grade 12 problems with charge, distance, direction, and energy

What you will learn

1. Set up the system and model

2. Electric field and force

3. Electric potential energy and potential

4. A reliable solution and checking method

Force between two charges

Field and force on a negative charge

Potential and potential energy

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review