HS-PS2-4: Newton's Law of Gravitation and Coulomb's Law
In plain English: HS-PS2-4 is the NGSS high school performance expectation that asks students to use Newton's law of gravitation and Coulomb's law to describe and predict the gravitational and electrostatic forces between two objects. Students calculate forces, predict how they change with mass, charge and distance, and describe the fields around objects. It is usually taught in Physics.
Use mathematical representations of Newton's Law of Gravitation and Coulomb's Law to describe and predict the gravitational and electrostatic forces between objects.
Clarification Statement: Emphasis is on both quantitative and conceptual descriptions of gravitational and electric fields.
Assessment Boundary: Assessment is limited to systems with two objects.
Next Generation Science Standards · Disciplinary Core Idea: Motion and Stability: Forces and Interactions · Official standard
Students use two inverse-square laws as mathematical models. Newton's law of gravitation, F = Gm1m2/r², gives the attractive force between two masses. Coulomb's law, F = kq1q2/r², gives the force between two charges, which is attractive for unlike charges and repulsive for like charges. Students calculate forces, predict how they change when a mass, a charge or the distance changes, analyze data to test the inverse-square pattern, and describe the gravitational and electric fields that surround objects: g = GM/r² in N/kg and E = kq/r² in N/C.
The NGSS Clarification Statement says: "Emphasis is on both quantitative and conceptual descriptions of gravitational and electric fields." The Assessment Boundary says: "Assessment is limited to systems with two objects." Every problem on this page involves exactly two objects, with r measured center to center. Constants used: G = 6.67 × 10⁻¹¹ N·m²/kg², k = 8.99 × 10⁹ N·m²/C², elementary charge e = 1.60 × 10⁻¹⁹ C, Earth's mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. Data sets are invented for teaching. Students do the practice of using mathematical representations: they compute, compare, graph and use the equations to make predictions they can check.
Learning Objectives
By the end of this lesson, students will be able to:
Calculate the gravitational force between two masses with F = Gm1m2/r² and the electrostatic force between two charges with F = kq1q2/r², using scientific notation and SI units
Predict how each force changes when a mass, a charge or the distance changes, using proportional reasoning
Describe the direction of each force (gravity always attracts; like charges repel and unlike charges attract) and explain why the two forces in a pair are equal and opposite
Describe gravitational and electric fields quantitatively (g = GM/r², E = kq/r²) and conceptually as the force per unit mass or charge that another object would feel
Use data to test whether a force follows an inverse-square relationship and compare the sizes of gravitational and electric forces
Prior Knowledge Required
Students should already be comfortable with:
Factors that affect the strength of electric forces MS-PS2-3
Gravitational interactions are attractive and depend on mass MS-PS2-4
Operations with numbers in scientific notation 8.EE.A.4
Rub a strip of clear tape off the table and hold it near a second strip: it bends away. Then remind students that Earth pulls on everything in the room. Ask:
Warm-Up Prompt
"You and the person next to you both have mass, so gravity pulls you toward each other. Why don't you drift together? The tape strips have almost no mass, yet they push each other visibly. What does that suggest about the sizes of the two kinds of force?"
Collect ideas. Students often say gravity needs a big object; ask what exactly makes Earth's pull large (its mass, not its size alone). Keep the question "how big is each force?" on the board to answer with numbers.
Direct Instruction15-20 minutes
Two inverse-square laws. Newton's law of gravitation: F = Gm1m2/r². Coulomb's law: F = kq1q2/r². In both, the force is proportional to the product of the two masses or charges and inversely proportional to the square of the distance between the centers. Doubling the distance gives one quarter of the force; tripling it gives one ninth. Gravity always attracts. Electric forces attract for unlike charges and repel for like charges. In each pair the two forces are equal in size and opposite in direction (Diagram 2). A field describes what one object does to the space around it: Earth's gravitational field strength is g = GM/r² (N/kg), and a charge's electric field is E = kq/r² (N/C). A second object placed in the field feels F = mg or F = qE.
Gravity between two people
Two students with masses 70 kg and 60 kg stand 1.0 m apart (center to center).
Result: F = (6.67 × 10⁻¹¹)(70)(60)/(1.0)² = 2.8 × 10⁻⁷ N, attractive. That is far too small to notice against friction from the floor.
Gravity between Earth and the Moon
Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg) are 3.84 × 10⁸ m apart.
Result: F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²)/(3.84 × 10⁸)² = 1.98 × 10²⁰ N. Earth pulls the Moon, and the Moon pulls Earth, with this same force.
Electrostatic force between two charged spheres
Two metal spheres, each 3.0 cm across, carry +50 nC (5.0 × 10⁻⁸ C) each, with centers 5.0 cm apart.
Result: F = (8.99 × 10⁹)(5.0 × 10⁻⁸)²/(0.050)² = 8.99 × 10⁻³ N, repulsive because the charges are alike.
Predicting with the inverse square
The same two spheres are moved from 5.0 cm to 15 cm apart. Predict the new force without recomputing from scratch.
Result: The distance is 3 times larger, so the force is 1/3² = 1/9 as large: 8.99 × 10⁻³ ÷ 9 ≈ 1.0 × 10⁻³ N.
Comparing the two forces in a hydrogen atom
A proton (1.67 × 10⁻²⁷ kg) and an electron (9.11 × 10⁻³¹ kg) are 5.29 × 10⁻¹¹ m apart, each with charge of size 1.60 × 10⁻¹⁹ C.
Result: Electric: 8.22 × 10⁻⁸ N. Gravitational: 3.63 × 10⁻⁴⁷ N. The electric force is about 2.3 × 10³⁹ times larger, and because both forces follow 1/r², the ratio does not depend on the distance.
Then show Diagram 1: Earth's field strength is 9.81 N/kg at the surface, where g = GM/R² = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)/(6.37 × 10⁶)², and it falls as 1/r² farther out. After each example ask: "Which quantity did you change, and by what factor should the force change?" Stress that r is measured between centers and that charges in microcoulombs or nanocoulombs must be converted to coulombs.
Guided Practice15-20 minutes
Pairs analyze invented data from a demonstration in which two identical charged metal spheres, each 3.0 cm across, are held at different distances and a sensitive force sensor measures the repulsion. They test whether the data follow Coulomb's law.
Guided practice data: force between two identical charged spheres
r (cm)
4.0
6.0
8.0
10.0
12.0
F (mN)
9.1
3.98
2.26
1.43
1.00
Compute F × r² for each column (in mN·cm²). The values stay close to 144, while F × r does not stay constant. What does that tell you about the relationship?
Plot F against 1/r². A straight line through the origin supports F = kq²/r².
Convert the constant to SI units: 144 mN·cm² = 1.44 × 10⁻⁵ N·m². Solve kq² = 1.44 × 10⁻⁵ for the charge on each sphere (q = 4.0 × 10⁻⁸ C, or 40 nC).
Describe the electric field of one sphere at 6.0 cm in words and numbers: E = kq/r², directed away from the sphere, and explain how it relates to the force on the other sphere.
Independent Practice10-15 minutes
Each student predicts the gravitational force on a 1,000 kg satellite at an altitude of one Earth radius above the surface, so r = 2R = 1.274 × 10⁷ m from Earth's center. They compute it directly with Newton's law (about 2,450 N), then check it with the inverse square (the surface weight of 9,810 N divided by 4). They describe Earth's field at that height (2.45 N/kg, pointing toward Earth's center) and write two sentences explaining why the satellite still has a large gravitational force on it even though astronauts in orbit seem weightless.
Closure5 minutes
Exit ticket: "Two charges attract each other with a force of 12 N. One charge is doubled and the distance between them is halved. Predict the new force and state its direction." (96 N, still attractive.) Then return to the warm-up and ask students to use Example 5 to explain why the tape strips visibly push each other while two students do not visibly attract.
Differentiation Strategies
For Struggling Students
Give a scaling table (factor on m1, factor on m2 or q, factor on r, factor on F) and practice proportional predictions before any calculator work
Provide a scientific-notation checklist: convert units first, multiply the leading numbers, add the exponents, then divide by r²
Use Diagram 1 to read field values at whole numbers of Earth radii before computing any in-between values
For Advanced Students
Ask students to derive the ratio of electric to gravitational force for two protons and explain why it does not depend on distance
Have students use g = GM/R² to estimate Earth's mass from the measured value of g and Earth's radius
Ask students to explain why a planet is nearly electrically neutral and what would happen to the Earth-Moon interaction if each carried a large net charge of the same sign
Assessment Guidance
What to Look For
Check that students square the distance, convert μC, nC and cm to SI units, and give a direction with every force. Look for predictions made by scaling ("three times as far, one ninth the force") and then confirmed by calculation. Listen for students who think the larger object pulls harder on the smaller one, who confuse field with force, or who treat gravity as absent in orbit.
02
Classroom Activities
3 Activities
1
Coulomb's Law Simulation Lab
25 minPairs
Pairs use a free online Coulomb's law simulation that shows two point charges on a line and the force between them. In Part A they change only the distance; in Part B they change only one charge. They make tables and graphs and decide what relationship the data support.
Procedure
Part A: set q1 = +3.0 μC and q2 = -1.0 μC and record the force at 0.10, 0.20, 0.30 and 0.40 m.
Part B: set r = 0.20 m and q2 = -1.0 μC, and record the force for q1 = +1.0, +2.0, +3.0 and +4.0 μC.
Graph F against r, F against 1/r² and F against q1.
Sample Data (invented)
Part A: F = 2.70, 0.674, 0.300 and 0.169 N (attractive). Part B: F = 0.225, 0.450, 0.674 and 0.899 N (attractive).
Discussion Questions
When the distance doubles from 0.10 m to 0.20 m, the force falls to one quarter. Show this from the data, and predict the force at 0.50 m before checking it in the simulation.
In Part A, F × r² stays at about 0.027 N·m². What combination of constants and charges does that number equal?
In Part B, what happens to the force each time q1 goes up by 1.0 μC? Which graph is a straight line through the origin?
Modification for Distance Learning
Students run the simulation at home and share their three graphs and a claim about each relationship on the class discussion board.
2
Sticky Tape Electrostatics
20 minPairs
Pairs charge strips of clear tape by pulling them apart and use them to describe electric forces qualitatively: which way each force points, how it changes with distance, and how it compares with gravity between the same two strips.
Procedure
Stick a 15 cm strip of tape on the table (the B strip) and another on top of it (the T strip). Fold a small tab at one end of each. Pull the pair off the table, then pull the T strip off the B strip quickly.
Hang each strip from a pencil and bring strips close: T near T, B near B, and T near B. Record attract or repel and how the bending changes as the strips get closer.
Keep strips away from metal and hands between tests, which can remove charge.
Sample Observations and Estimate (invented)
T and T repel, B and B repel, T and B attract, and every effect gets stronger as the strips get closer. Each strip has a mass of about 0.20 g. Estimated gravitational force between two strips 1.0 cm apart: (6.67 × 10⁻¹¹)(2.0 × 10⁻⁴)²/(0.010)² ≈ 2.7 × 10⁻¹⁴ N. Weight of one strip: about 2.0 × 10⁻³ N.
Discussion Questions
What do the observations tell you about the charges on T and B strips?
The gravitational force between two strips is less than one billionth of a strip's weight. How can you tell that the bending you see must come from electric forces?
Which parts of Coulomb's law can you describe from these observations alone, and which need measurements?
3
Force Comparison Stations
25 minGroups of 3-4
Four stations each describe two objects. Groups compute the force or field, state its direction, and rank the stations. Groups rotate every 6 minutes.
Station Cards
Station 1, bowling balls: two 7.0 kg balls 0.30 m apart, center to center. Find the gravitational force.
Station 2, Sun and Earth: Sun 1.99 × 10³⁰ kg, Earth 5.97 × 10²⁴ kg, 1.50 × 10¹¹ m apart. Find the gravitational force.
Station 3, two protons: two protons 3.0 × 10⁻¹⁵ m apart, about the size of a small nucleus. Find the electric force. (What holds a nucleus together anyway is beyond this standard.)
Station 4, Mars: mass 6.42 × 10²³ kg, radius 3.39 × 10⁶ m. Find the gravitational field strength at the surface and the weight of a 20 kg rover there.
Answer Key for Teachers
Station 1: 3.6 × 10⁻⁸ N. Station 2: 3.52 × 10²² N. Station 3: about 26 N, repulsive. Station 4: 3.73 N/kg, so the rover weighs about 75 N.
Discussion Questions
Station 2 has by far the largest force. Which factor in the equation makes it so large?
The two protons have tiny masses, yet their electric force is roughly 700 million times the gravitational force between the bowling balls. What does this say about the relative strength of the two interactions?
At Station 4, how would the field strength change on a planet with the same mass and twice the radius?
Modification for Distance Learning
Post each station on a slide. Groups work in breakout rooms and submit their four calculations and a ranking with one sentence of reasoning.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Earth's Gravitational Field Falls Off as 1/r²
Earth's gravitational field strength g = GM/r², drawn to scale from the surface (1 Earth radius) out to 4.5 Earth radii. At 2, 3 and 4 Earth radii from the center, g is one quarter, one ninth and one sixteenth of its surface value of 9.81 N/kg. A mass m placed at any point feels a force F = mg toward Earth's center.
Diagram 2: Force Pairs for Gravity and Electric Charge
Gravity between two masses is always attractive. Electric forces repel for like charges and attract for unlike charges. In each pair the two forces are equal in size and opposite in direction, even when one object is much larger, and r is always measured from center to center.
04
Homework Assignment
~30 min
HS-PS2-4 Homework: Predicting Gravitational and Electric Forces
Directions: Every problem involves two objects. Use G = 6.67 × 10⁻¹¹ N·m²/kg², k = 8.99 × 10⁹ N·m²/C², Earth's mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. Convert all units to SI, give answers to three significant figures, and state the direction of every force.
Part 1: Calculating Forces (Problems 1-2)
A 75 kg astronaut is on a space station orbiting 420 km above Earth's surface. (a) Find her distance from Earth's center. (b) Find the gravitational force Earth exerts on her. (c) Compare it with her weight on Earth's surface, as a percent. (d) Use your answer to respond to the claim "there is no gravity in space."
A metal sphere 4.0 cm across with charge +40 nC is 0.30 m (center to center) from an identical sphere with charge -60 nC. (a) Find the size of the electric force and say whether it is attractive or repulsive. (b) Predict the force if the spheres are moved to 0.60 m apart, and check your prediction with Coulomb's law. (c) Describe the direction of the force on each sphere.
Part 2: Data and Fields (Problems 3-4)
A student measures the force between two charged objects at three distances: r = 0.10, 0.20 and 0.40 m gives F = 8.0, 2.0 and 0.50 mN. (a) Test whether the data fit F ∝ 1/r or F ∝ 1/r², showing your work. (b) Predict the force at 0.25 m. (c) Predict the force at 0.10 m if one of the charges is doubled.
The Moon has a mass of 7.35 × 10²² kg and a radius of 1.74 × 10⁶ m. (a) Find the gravitational field strength at its surface. (b) Find the weight of a 70 kg astronaut on the Moon and on Earth (g = 9.81 N/kg). (c) Describe in words what the Moon's field would do to a small mass placed near its surface.
Part 3: Fields and Comparisons (Problems 5-6)
A metal sphere 20 cm across carries +2.0 μC. (a) Find the size and direction of its electric field 0.50 m away. (b) A -1.0 nC charge is placed at that point. Find the force on it from F = qE, and state its direction. (c) Check your answer to (b) with Coulomb's law.
Two hollow metal spheres, each 1.0 kg and 20 cm across, are 1.0 m apart (center to center), and each carries an extra charge of +2.0 μC. (a) Find the gravitational force between them. (b) Find the electric force between them and its direction. (c) Find the ratio of the two forces. (d) Electric forces are so much stronger, yet gravity controls the motion of planets and moons. Write a short argument explaining why.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Calculations
Correct use of both laws with SI units, squared distances and correct scientific notation
Correct method with one unit, exponent or squaring error
Predictions
Predictions made by scaling and confirmed by calculation, with the factor stated
Prediction given without reasoning, or scaling factor wrong
Descriptions of Forces and Fields
States the direction of every force and describes fields as force per unit mass or charge
Directions or field descriptions incomplete
Using Data and Arguments
Tests the inverse-square pattern with the data and writes an argument backed by numbers
Pattern named or argument made without using the data
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Every question involves two objects. Use G = 6.67 × 10⁻¹¹ N·m²/kg² and k = 8.99 × 10⁹ N·m²/C². Data are invented. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
The distance between the centers of two asteroids is cut in half. What happens to the gravitational force between them?
Answer: C
F is proportional to 1/r², so halving r multiplies F by 2² = 4. Choice A treats the law as 1/r instead of 1/r². Choice B reverses the relationship.
Question 2 of 20 · Multiple Choice
One of two charges is tripled while the distance stays the same. How does the electric force change?
Answer: A
F = kq1q2/r² is proportional to each charge, so tripling one charge triples the force. Choice B squares the factor, which only applies to the distance. Choice C inverts the relationship.
Question 3 of 20 · Multiple Choice
Two 500 kg boulders are 2.0 m apart, center to center. What is the gravitational force between them?
Answer: D
F = (6.67 × 10⁻¹¹)(500)(500)/(2.0)² = (1.67 × 10⁻⁵)/4 = 4.17 × 10⁻⁶ N. Choice A divides by 2.0 instead of 2.0². Choice B leaves out the distance entirely. Choice C drops the negative sign of the exponent.
Question 4 of 20 · Multiple Choice
Two metal spheres, each 3.0 cm across, carry +20 nC and +30 nC, with centers 0.20 m apart. What is the electric force between them?
Answer: B
F = (8.99 × 10⁹)(2.0 × 10⁻⁸)(3.0 × 10⁻⁸)/(0.20)² = 1.35 × 10⁻⁴ N, and like charges repel. Choice A has the right size but the wrong direction. Choice C divides by 0.20 instead of 0.20². Choice D forgets to convert nanocoulombs to coulombs.
Question 5 of 20 · Multiple Choice
A positively charged rod is brought near a small negatively charged bead. What do they do?
Answer: A
Unlike charges attract. Choice C forgets Newton's third law: the rod feels an equal and opposite force. Choice D ignores that electric forces act at a distance through the field.
Question 6 of 20 · Multiple Choice
An apple hangs from a tree. How does the gravitational force of Earth on the apple compare with the gravitational force of the apple on Earth?
Answer: C
Newton's law of gravitation uses the product m1m2, which is the same for both objects, and Newton's third law says the forces are equal and opposite. The effects differ: the apple accelerates noticeably and Earth does not, because Earth's mass is enormous. Choice A confuses force with acceleration.
Question 7 of 20 · Multiple Choice
Earth's gravitational field strength is 9.81 N/kg at the surface. What is it at a distance of 5 Earth radii from Earth's center?
Answer: D
g = GM/r², so at 5 times the distance g is 1/25 as large: 9.81 ÷ 25 = 0.392 N/kg. Choice A divides by 5 instead of 25. Choice B divides by 5³ = 125, as if the field fell off as 1/r³. Choice C assumes the field is the same everywhere.
Question 8 of 20 · Multiple Choice
Four data sets give a force at r = 1, 2 and 3 units. Which one follows an inverse-square law?
Answer: B
For an inverse-square law, F × r² is constant: 36 × 1 = 9 × 4 = 4 × 9 = 36. Choice A keeps F × r constant (12 × 1 = 6 × 2 = 4 × 3), which is an inverse relationship, not an inverse square. Choice C increases with distance, and choice D falls by a constant amount.
Question 9 of 20 · Multiple Choice
What does the electric field at a point near a charged object describe?
Answer: A
E = F/q: the field tells you the force per coulomb that any charge placed there would feel, with the direction of the force on a positive charge. Choice B confuses the field with its source. Choice C is impossible: an object does not exert a net force on itself.
Question 10 of 20 · Multiple Choice
The electric force between two charged particles is far stronger than gravity for the same distance. Why does gravity, not electric force, control how the Moon moves around Earth?
Answer: C
Each body contains nearly equal amounts of positive and negative charge, so the net charge, and therefore the electric force, is tiny. Gravity has only one sign, so it adds up over huge masses. Choice B is false: both laws are inverse-square. Choice A is false: electric forces reach just as far, with the same 1/r² fall-off.
Question 11 of 20 · Multiple Choice
Two charges exert a force of 18 N on each other. The distance between them is tripled. What is the new force?
Answer: D
Tripling r divides F by 3² = 9: 18 ÷ 9 = 2.0 N. Choice A divides by 3, using 1/r. Choice B multiplies by 3, and choice C multiplies by 9: both reverse the relationship.
Question 12 of 20 · Multiple Choice
Both masses in a two-object system are doubled, and the distance between them is also doubled. What happens to the gravitational force?
Answer: B
Doubling both masses multiplies F by 2 × 2 = 4, and doubling r divides it by 2² = 4, so F is unchanged. Choice C forgets the distance change. Choice D forgets the mass changes.
Question 13 of 20 · Multiple Choice
Jupiter has a mass of 1.90 × 10²⁷ kg and a radius of 6.99 × 10⁷ m. What is the gravitational field strength at the top of its clouds?
Answer: A
g = GM/R² = (6.67 × 10⁻¹¹)(1.90 × 10²⁷)/(6.99 × 10⁷)² = 25.9 N/kg. Choice B divides by R instead of R². Choice C assumes every planet has Earth's field strength. Choice D is off by a factor of 10, an exponent slip.
Question 14 of 20 · Multiple Choice
A small negative charge is placed near a positively charged sphere. The sphere's electric field points away from the sphere. Which way is the force on the negative charge?
Answer: C
The field direction is the direction of the force on a positive charge. A negative charge feels a force opposite to the field, so it is pulled toward the positive sphere, as expected for unlike charges. Choice A describes a positive charge. Choice B is impossible: toward the sphere is opposite to a field that points away from it.
Question 15 of 20 · Short Answer
Use Newton's law of gravitation to find the force between Earth and a 5.0 kg bowling ball resting on the surface. Compare it with mg, using g = 9.81 N/kg, and explain the result.
F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(5.0)/(6.37 × 10⁶)² = 49.1 N. Also mg = 5.0 × 9.81 = 49.05 N, about 49.1 N. They agree because the weight mg is the gravitational force of Earth, and g is just GM/R², Earth's field strength at the surface.
Question 16 of 20 · Short Answer
Two metal spheres, each 4.0 cm across and each with charge +60 nC, repel each other with a force of 3.6 × 10⁻³ N. How far apart are their centers?
Rearrange F = kq²/r²: r = √(kq²/F) = √((8.99 × 10⁹)(6.0 × 10⁻⁸)²/(3.6 × 10⁻³)) = √0.00899 ≈ 0.0948 m (about 9.5 cm) between centers.
Question 17 of 20 · Short Answer
In an invented version of a classic experiment, a 158 kg lead sphere and a 0.730 kg lead sphere are held 0.230 m apart, center to center, and the measured attraction is 1.45 × 10⁻⁷ N. Use the data to find G, and compare it with the accepted value.
G = Fr²/(m1m2) = (1.45 × 10⁻⁷)(0.230)²/(158 × 0.730) = 6.65 × 10⁻¹¹ N·m²/kg², within about 0.3% of the accepted 6.67 × 10⁻¹¹ N·m²/kg², so the data support Newton's law of gravitation.
Question 18 of 20 · Short Answer
Two charges exert a force F on each other. Predict the new force if the first charge is halved, the second charge is multiplied by 4, and the distance is doubled. Show the factors.
Factor from charges: ½ × 4 = 2. Factor from distance: 1/2² = ¼. Overall: 2 × ¼ = ½, so the new force is F/2, in the same direction as before (the signs did not change).
Question 19 of 20 · Short Answer
Describe two ways Earth's gravitational field and the electric field of a small positively charged sphere are alike and two ways they are different.
Alike: both weaken as 1/r² with distance from the center, and both describe the force per unit (per kg or per C) that another object would feel there. Different: the gravitational field always pulls masses toward Earth, while the electric field pushes positive charges away and pulls negative charges toward the sphere; and the electric field depends on charge (in N/C) while the gravitational field depends on mass (in N/kg). For ordinary objects the electric force per particle is also far stronger.
Question 20 of 20 · Short Answer
For two electrons, find the ratio of the electric force to the gravitational force (electron mass 9.11 × 10⁻³¹ kg, charge size 1.60 × 10⁻¹⁹ C). Explain why you do not need to know the distance between them.
Ratio = ke²/(Gm²) = (8.99 × 10⁹)(1.60 × 10⁻¹⁹)² ÷ [(6.67 × 10⁻¹¹)(9.11 × 10⁻³¹)²] ≈ 4.16 × 10⁴². Both forces are proportional to 1/r², so r² cancels in the ratio: the electric repulsion is that many times stronger than the gravitational attraction at any distance.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-PS2-4 mean?
HS-PS2-4 asks students to use the equations for Newton's law of gravitation and Coulomb's law to describe and predict gravitational and electric forces between objects. Students calculate forces, predict how they change with mass, charge and distance, and describe the fields around objects. The code means High School, Physical Science, core idea 2 (Motion and Stability: Forces and Interactions), performance expectation 4.
What is the assessment boundary for HS-PS2-4?
Assessment is limited to systems with two objects. Students work with one pair of masses or one pair of charges at a time. Adding the forces from three or more objects (superposition, often with vectors in two dimensions) is beyond this standard.
Is HS-PS2-4 taught in physics or chemistry?
It is usually taught in high school Physics, in the forces unit or a unit on fields. Chemistry courses use Coulomb's law qualitatively to explain attractions between ions and electrons, but the calculations in this standard fit a physics course. The math needed is scientific notation and proportional reasoning.
How are Newton's law of gravitation and Coulomb's law alike?
Both are inverse-square laws: the force is proportional to the product of two quantities (masses or charges) and to 1/r². Both act at a distance, both give equal and opposite forces on the two objects, and both can be described with a field. The differences are the constant, the property (mass or charge), and the fact that gravity only attracts while electric forces can attract or repel.
What is the difference between a force and a field?
A force acts between two objects; a field belongs to one object and describes the force per unit mass or charge that another object would feel at each point. Earth's gravitational field at the surface is about 9.81 N/kg, so a 2 kg object there feels about 19.6 N. The Clarification Statement asks for both quantitative and conceptual descriptions of fields.
Why is gravity so weak compared with the electric force?
Because G is extremely small compared with k, so for single particles the gravitational force is smaller by a huge factor. That is why electric forces control atoms, molecules and the charged tape strips in this lesson, while gravity between everyday objects is too small to notice.
What are common mistakes on HS-PS2-4 problems?
A common one is forgetting to square the distance. Others are leaving charges in μC or nC, measuring r from surfaces instead of centers, and giving a force without its direction. Many students also think the larger object pulls harder on the smaller one, when the two forces are always equal and opposite.
Do students need vectors for HS-PS2-4?
Only direction along the line between the two objects: toward or away from the other object. Because the standard is limited to two objects, there is no need to add force vectors at angles. Signs and words (attractive, repulsive) are enough.
What labs work for gravitation and Coulomb's law?
Gravitational forces between classroom objects are too small to measure without special equipment, so students usually analyze data from simulations or from published experiments. Electric forces are easier to see: charged tape strips and balloons show attraction, repulsion and the effect of distance, and Coulomb's law simulations let students collect force-distance data quickly.
What comes after HS-PS2-4?
HS-PS3-5 asks students to model two objects interacting through electric or magnetic fields and the energy changes involved, and HS-ESS1-4 uses gravitation to predict the motion of orbiting objects. HS-PS2-5 then connects electric currents and magnetic fields.
07
Related Standards
5 standards
These standards connect to HS-PS2-4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-PS2-4Prerequisite
Argue that gravitational interactions are attractive and depend on the masses
NGSS is a registered trademark of WestEd. Neither WestEd nor the lead states and partners that developed the Next Generation Science Standards were involved in the production of this product, and do not endorse it.
