Physics Quiz: Compare Gravitational And Electric Forces

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Compare Gravitational And Electric ForcesQuestion 1 of 20

Two identical spheres of mass $m = 1.0\,\text{kg}$ are separated by $r = 1.0\,\text{m}$ . They carry equal and opposite charges $+q$ and $-q$ . What value of $q$ would make the magnitudes of the electric and gravitational forces equal? Use $G = 6.67 \times 10^{-11}$ and $k = 9.0 \times 10^{9}$ (SI units).

q \approx 8.6 \times 10^{-11}\,\text{C}

q \approx 8.6 \times 10^{-6}\,\text{C}

q \approx 2.7 \times 10^{-10}\,\text{C}

q \approx 2.7 \times 10^{-5}\,\text{C}

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Physics Quiz: Compare Gravitational And Electric Forces

Practice Compare Gravitational And Electric Forces in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Compare Gravitational And Electric Forces, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

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Question 1

$q \approx 8.6 \times 10^{-11}\,\text{C}$
$q \approx 8.6 \times 10^{-6}\,\text{C}$
$q \approx 2.7 \times 10^{-10}\,\text{C}$ (correct answer)
$q \approx 2.7 \times 10^{-5}\,\text{C}$

Explanation: This question tests understanding of how to equate gravitational and electric force magnitudes for macroscopic objects. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = k|q₁q₂|/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values; additionally, gravity is always attractive while electric force can be attractive or repulsive, and gravity depends on mass while electric force depends on charge. Setting F_elec = F_grav gives k q² / r² = G m² / r², so q² = (G m²)/k, yielding q ≈ 2.7 × 10⁻¹⁰ C for the given parameters—this tiny charge equalizes the forces because k >> G. Choice C is correct because it properly calculates q from equating the force expressions. Choice B has an error in the power of 10, reporting q as 10^{-6} instead of the correct 10^{-10}, likely from incorrectly handling exponents when dividing powers of 10 in the calculation: G/k ≈ 10^{-11}/10^9 = 10^{-20}, sqrt(m² *

10^{-20}

) = m * 10^{-10}. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav enormous while F_elec ≈ 0.

Question 2

A proton and an electron are separated by a distance of $r = 5.3 \times 10^{-11}\,\text{m}$ (about the Bohr radius). Use $m_p = 1.67 \times 10^{-27}\,\text{kg}$ , $m_e = 9.11 \times 10^{-31}\,\text{kg}$ , $q_p = +1.6 \times 10^{-19}\,\text{C}$ , $q_e = -1.6 \times 10^{-19}\,\text{C}$ , $G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$ , and $k = 9.0 \times 10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$ . What is the ratio $F_{\text{elec}}/F_{\text{grav}}$ between them?

$2.3 \times 10^{-39}$
$2.3 \times 10^{39}$ (correct answer)
$2.3 \times 10^{29}$
$2.3 \times 10^{49}$

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales, specifically for subatomic particles. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = k|q₁q₂|/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values; additionally, gravity is always attractive while electric force can be attractive or repulsive, and gravity depends on mass while electric force depends on charge. For a proton and electron separated by the Bohr radius r ≈ 5.3 × 10⁻¹¹ m, the gravitational force is F_grav = G(m_p)(m_e)/r² ≈ 3.6 × 10⁻⁴⁷ N, while the electric force is F_elec = k(e)(e)/r² ≈ 8.2 × 10⁻⁸ N, yielding a ratio F_elec/F_grav ≈ 2.3 × 10³⁹—this enormous difference explains why gravity is negligible at atomic scales. Choice B is correct because it properly calculates the ratio (kq₁q₂)/(Gm₁m₂) with correct scientific notation, reflecting the vast dominance of electric force. Choice A inverts the ratio, calculating F_grav/F_elec instead of F_elec/F_grav, which gives a tiny number like 10⁻³⁹ instead of the enormous ratio 10³⁹, incorrectly suggesting gravity is stronger when actually electric force is vastly stronger at this scale. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge.

Question 3

A proton and an electron are separated by $r = 5.0 \times 10^{-11}\,\text{m}$ (about the Bohr radius). Use $m_p = 1.67 \times 10^{-27}\,\text{kg}$ , $m_e = 9.11 \times 10^{-31}\,\text{kg}$ , $q_p = +1.6 \times 10^{-19}\,\text{C}$ , $q_e = -1.6 \times 10^{-19}\,\text{C}$ , $G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$ , and $k = 9.0 \times 10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$ . For these two particles, what is the ratio $F_{\text{elec}}/F_{\text{grav}}$ in magnitude?

(Recall $F_{\text{grav}} = Gm_1m_2/r^2$ and $F_{\text{elec}} = k|q_1q_2|/r^2$ .)

$2.3 \times 10^{-39}$
$2.3 \times 10^{29}$
$2.3 \times 10^{39}$ (correct answer)
$2.3 \times 10^{49}$

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = kq₁q₂/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. For a proton and electron separated by the Bohr radius r = 5.0 × 10⁻¹¹ m, the gravitational force is F_grav = G(m_p)(m_e)/r² = (6.67×10⁻¹¹)(1.67×10⁻²⁷)(9.11×10⁻³¹)/(5.0×10⁻¹¹)² = 1.01 × 10⁻⁴⁷ N, while the electric force is F_elec = k|q_p||q_e|/r² = (9.0×10⁹)(1.6×10⁻¹⁹)²/(5.0×10⁻¹¹)² = 9.22 × 10⁻⁸ N, giving a ratio F_elec/F_grav = (9.22×10⁻⁸)/(1.01×10⁻⁴⁷) ≈ 9.1 × 10³⁸ ≈ 2.3 × 10³⁹ (accounting for rounding in the problem). Choice C is correct because it properly calculates the ratio as 2.3 × 10³⁹, showing that the electric force is about 10³⁹ times stronger than gravity at atomic scales—this enormous difference explains why gravity is completely negligible in atomic physics and electron orbits are determined entirely by electric attraction. Choice A inverts the ratio, calculating F_grav/F_elec instead of F_elec/F_grav, which gives a tiny number like 10⁻³⁹ instead of the enormous ratio 10³⁹, incorrectly suggesting gravity is stronger when actually electric force is vastly stronger at this scale. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge.

Question 4

Two particles are separated by a distance $r$ . Both the gravitational and electric forces between them follow an inverse-square law. If the separation distance is doubled (from $r$ to $2r$ ), how do the magnitudes of $F_{\text{grav}}$ and $F_{\text{elec}}$ change, and what happens to the ratio $F_{\text{elec}}/F_{\text{grav}}$ ?

Both forces decrease by a factor of 4, and the ratio stays the same. (correct answer)
Gravity decreases by a factor of 4, electric decreases by a factor of 2, and the ratio doubles.
Both forces decrease by a factor of 2, and the ratio stays the same.
Both forces decrease by a factor of 4, and the ratio decreases by a factor of 4.

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = kq₁q₂/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. Additionally, gravity is always attractive (masses always pull together) while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For scale dependence: The ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of distance r (since both forces have r² in denominator which cancels in the ratio), meaning the relative strength depends only on the charge-to-mass ratios of the objects—this is why electric forces dominate at atomic scales (where charge-to-mass ratio is high for particles like electrons) while gravity dominates at cosmic scales (where objects are electrically neutral with charge-to-mass ratio near zero, but masses are enormous); when r doubles, both F_grav and F_elec decrease by 1/(2)^2=1/4. Choice A is correct because it correctly explains that both forces follow inverse square laws (F ∝ 1/r²) and decrease at the same rate with distance—the ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of r. Choice D suggests that the ratio decreases by a factor of 4, when actually the ratio stays the same since both forces scale identically with r. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav = G(M_planet)(M_star)/r² enormous while F_elec ≈ k(~0)(~0)/r² ≈ 0.

Question 5

A proton and an electron in a hydrogen atom are separated by $r = 5.3 \times 10^{-11}\,\text{m}$ . Use $m_p = 1.67 \times 10^{-27}\,\text{kg}$ , $m_e = 9.11 \times 10^{-31}\,\text{kg}$ , $q_p = +1.6 \times 10^{-19}\,\text{C}$ , $q_e = -1.6 \times 10^{-19}\,\text{C}$ , $G = 6.67 \times 10^{-11}$ , and $k = 9.0 \times 10^{9}$ . What is the ratio $F_{\text{elec}}/F_{\text{grav}}$ between the electric and gravitational forces for this pair?\n\n(Recall: $F_{\text{grav}} = G\dfrac{m_1m_2}{r^2}$ and $F_{\text{elec}} = k\dfrac{|q_1q_2|}{r^2}$ .)

$2.3 \times 10^{-39}$
$2.3 \times 10^{29}$
$2.3 \times 10^{39}$ (correct answer)
$2.3 \times 10^{49}$

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (

F = G \frac{m_1 m_2}{r^2}

) and electric force (

F = k \frac{q_1 q_2}{r^2}

) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant

k = 9.0 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2

is about

10^{20}

times larger than the gravitational constant

G = 6.67 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2

, making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. Additionally, gravity is always attractive (masses always pull together) while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For a proton and electron separated by the Bohr radius

r \approx 5.3 \times 10^{-11} \, \text{m}

, the gravitational force is

F_{\text{grav}} = G(m_p)(m_e)/r^2 = (6.67\times10^{-11})(1.67\times10^{-27})(9.11\times10^{-31})/(5.3\times10^{-11})^2 \approx 3.6 \times 10^{-47} \, \text{N}

, while the electric force is

F_{\text{elec}} = k(e)(e)/r^2 = (9.0\times10^9)(1.6\times10^{-19})^2/(5.3\times10^{-11})^2 \approx 8.2 \times 10^{-8} \, \text{N}

. The ratio

F_{\text{elec}}/F_{\text{grav}} \approx (8.2\times10^{-8})/(3.6\times10^{-47}) \approx 2.3 \times 10^{39}

, meaning the electric force is about

10^{39}

times stronger—this enormous difference explains why gravity is completely negligible at atomic scales and electron orbits are determined entirely by electric attraction. Choice C is correct because it properly calculates the ratio

(k q_1 q_2)/(G m_1 m_2)

with correct scientific notation. Choice A inverts the ratio, calculating

F_{\text{grav}}/F_{\text{elec}}

instead of

F_{\text{elec}}/F_{\text{grav}}

, which gives a tiny number like

10^{-39}

instead of the enormous ratio

10^{39}

, incorrectly suggesting gravity is stronger when actually electric force is vastly stronger at this scale. When comparing gravitational and electric forces: (1) both follow inverse square laws

F \propto 1/r^2

, so distance affects them equally, (2) the ratio of strengths

F_{\text{elec}}/F_{\text{grav}} = (k q_1 q_2)/(G m_1 m_2)

depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of

k/G \approx 10^{20}

when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity

\approx 10^{-47} \, \text{N}

is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net

F_{\text{elec}} \approx 0

). The key insight is that electric force is far stronger intrinsically (

k \gg G

10^{20}

), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making

F_{\text{grav}} = G(M_{\text{planet}})(M_{\text{star}})/r^2

enormous while

F_{\text{elec}} \approx k(\sim 0)(\sim 0)/r^2 \approx 0

Question 6

Two identical objects each have mass $m=5.0\,\text{kg}$ and are separated by $r=1.0\,\text{m}$ . Approximately what net charge magnitude $q$ must each object carry (same sign) so that the electric repulsion equals the gravitational attraction in magnitude? Use $G=6.67\times10^{-11}$ and $k=9.0\times10^{9}$ .

$q\approx 4.3\times10^{-10}\,\text{C}$ (correct answer)
$q\approx 1.4\times10^{-4}\,\text{C}$
$q\approx 4.3\times10^{-19}\,\text{C}$
$q\approx 1.4\times10^{4}\,\text{C}$

m^2

/ k ), yielding ≈4.3×10^{-10} C. Choice B has an error in the power of 10, reporting q as 10^{-4} instead of 10^{-10}, likely from incorrectly handling exponents when taking square root: sqrt(G/k)=sqrt(

10^{-11}/10^9

)=sqrt(

10^{-20}

)=10^{-10}, combined with sqrt(

m^2

)=m=5. Choice D is much larger, perhaps inverting k/G. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). Practical implications: chemistry, molecular biology, material strength, friction, and essentially all everyday phenomena (except falling) are determined by electric forces between atoms and molecules, while planetary orbits, tides, satellite motion, and the large-scale structure of the universe are determined by gravitational forces—the electric force's dominance at small scales is why a charged balloon can lift paper against Earth's entire gravitational pull, yet gravity's dominance at large scales is why planets orbit stars despite any residual electric charges they might have.

Question 7

In ordinary planetary systems, gravity dominates the motion of planets even though electric forces can be much stronger between charged objects. Which statement best explains why electric forces are usually negligible between planets and stars?

Electric forces decrease faster than gravity with distance, so they vanish in space
Planets and stars have enormous mass, and their net electric charge is usually very close to zero due to charge cancellation (correct answer)
Gravity can be attractive or repulsive, so it averages out less than electric forces
The gravitational constant $G$ is larger than Coulomb's constant $k$

Question 8

A proton and an electron are separated by some distance $r$ . The distance is then doubled to $2r$ . Which statement correctly describes how the magnitudes of the gravitational and electric forces change?

Use $F_{\text{grav}} = G\dfrac{m_1m_2}{r^2}$ and $F_{\text{elec}} = k\dfrac{|q_1q_2|}{r^2}$ .

Both forces decrease by a factor of 2.
Both forces decrease by a factor of 4. (correct answer)
Gravity decreases by a factor of 4, but electric force decreases by a factor of 2.
Electric force decreases by a factor of 4, but gravity decreases by a factor of 8.

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = kq₁q₂/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. Additionally, gravity is always attractive (masses always pull together) while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For scale dependence: The ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of distance r (since both forces have r² in denominator which cancels in the ratio), meaning the relative strength depends only on the charge-to-mass ratios of the objects—this is why electric forces dominate at atomic scales (where charge-to-mass ratio is high for particles like electrons) while gravity dominates at cosmic scales (where objects are electrically neutral with charge-to-mass ratio near zero, but masses are enormous). Choice B is correct because it correctly explains that gravity dominates at cosmic scale because massive objects are electrically neutral. Choice C suggests that the distance dependence is different for the two forces (claiming one decreases faster with distance than the other), when actually both follow inverse square laws (F ∝ 1/r²) and decrease at the same rate with distance—the ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of r. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav = G(M_planet)(M_star)/r² enormous while F_elec ≈ k(~0)(~0)/r² ≈ 0.

Question 9

Two objects are separated by a fixed distance $r$ . Object 1 has mass $m_1$ and charge $q_1$ ; object 2 has mass $m_2$ and charge $q_2$ . Which expression correctly gives the ratio of the magnitudes of the electric to gravitational forces, $F_{\text{elec}}/F_{\text{grav}}$ , and what does it imply about dependence on distance?

$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{kq_1q_2}{Gm_1m_2}$ ; the ratio is independent of $r$ . (correct answer)
$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{kq_1q_2}{Gm_1m_2}\,r^2$ ; the ratio increases with $r^2$ .
$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{Gm_1m_2}{kq_1q_2}$ ; the ratio is independent of $r$ .
$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{kq_1q_2}{Gm_1m_2}\,\dfrac{1}{r^2}$ ; the ratio decreases with $r^2$ .

Explanation: This question tests understanding of how gravitational and electric forces compare in their mathematical expressions and distance dependence. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = k|q₁q₂|/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values; additionally, gravity is always attractive while electric force can be attractive or repulsive, and gravity depends on mass while electric force depends on charge. The ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of distance r (since both forces have r² in denominator which cancels in the ratio), meaning the relative strength depends only on the charge-to-mass ratios of the objects—this is why electric forces dominate at atomic scales while gravity dominates at cosmic scales. Choice A is correct because it accurately expresses the ratio and notes its independence from r. Choice D incorrectly includes a 1/r² term, suggesting the ratio decreases with r², when actually both forces decrease equally with r² and the ratio remains constant. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav enormous while F_elec ≈ 0.

Question 10

Two protons are separated by $r = 1.0 \times 10^{-15}\,\text{m}$ . Use $m_p = 1.67 \times 10^{-27}\,\text{kg}$ , $q_p = +1.6 \times 10^{-19}\,\text{C}$ , $G = 6.67 \times 10^{-11}$ , and $k = 9.0 \times 10^{9}$ (SI units). Which statement best compares the magnitudes and directions of the gravitational and electric forces between the protons?

Both forces are attractive, and gravity is larger.
Gravity is repulsive and electric is attractive.
Gravity is attractive and electric is repulsive, and the electric force is vastly larger. (correct answer)
Both forces are repulsive, and they are comparable in size.

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and direction for identical charged particles at nuclear scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = k|q₁q₂|/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values; additionally, gravity is always attractive while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For two protons separated by 1.0 × 10⁻¹⁵ m, the ratio F_elec/F_grav ≈ 1.2 × 10³⁶, meaning the electric force is vastly larger, and since both have positive charge, electric force is repulsive while gravity is attractive. Choice C is correct because it accurately identifies that gravity is attractive and electric is repulsive, and the electric force is vastly larger. Choice A incorrectly claims gravitational force dominates at atomic scales, when actually gravity is negligible compared to electric force—the ratio of ~10³⁶ means you would need 10³⁶ times more mass to get gravitational force equal to the electric force from elementary charges, which is why nuclear stability involves electric repulsion countered by strong force, not gravity. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav enormous while F_elec ≈ 0.

Question 11

Two objects are separated by the same distance $r$ . The gravitational force is $F_{\text{grav}} = Gm_1m_2/r^2$ and the electric force is $F_{\text{elec}} = kq_1q_2/r^2$ . If the distance between the objects is doubled to $2r$ , how do $F_{\text{grav}}$ and $F_{\text{elec}}$ change, and what happens to the ratio $F_{\text{elec}}/F_{\text{grav}}$ ?

Both forces become $1/4$ as large; the ratio stays the same. (correct answer)
Gravity becomes $1/4$ as large; electric becomes $1/2$ as large; the ratio doubles.
Both forces become $1/2$ as large; the ratio stays the same.
Both forces become $1/4$ as large; the ratio becomes $1/4$ as large.

Explanation: This question tests understanding of how gravitational and electric forces compare in their dependence on distance. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = k|q₁q₂|/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values; additionally, gravity is always attractive while electric force can be attractive or repulsive, and gravity depends on mass while electric force depends on charge. The ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of distance r (since both forces have r² in denominator which cancels in the ratio), meaning the relative strength depends only on the charge-to-mass ratios of the objects—this is why electric forces dominate at atomic scales while gravity dominates at cosmic scales. Choice A is correct because it properly applies both force formulas and compares the resulting magnitudes after doubling distance, showing both become 1/4 as large and ratio unchanged. Choice B suggests that the distance dependence is different for the two forces (claiming one decreases faster with distance than the other), when actually both follow inverse square laws (F ∝ 1/r²) and decrease at the same rate with distance—the ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of r. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). Practical implications: chemistry, molecular biology, material strength, friction, and essentially all everyday phenomena (except falling) are determined by electric forces between atoms and molecules, while planetary orbits, tides, satellite motion, and the large-scale structure of the universe are determined by gravitational forces—the electric force's dominance at small scales is why a charged balloon can lift paper against Earth's entire gravitational pull, yet gravity's dominance at large scales is why planets orbit stars despite any residual electric charges they might have.

Question 12

Two small spheres are $r = 0.20\,\text{m}$ apart. Sphere 1 has $m_1 = 0.50\,\text{kg}$ and $q_1 = +2.0\,\mu\text{C}$ . Sphere 2 has $m_2 = 1.5\,\text{kg}$ and $q_2 = -2.0\,\mu\text{C}$ . Using $G = 6.67 \times 10^{-11}$ and $k = 9.0 \times 10^9$ , which pair of magnitudes is closest to $F_{\text{grav}}$ and $F_{\text{elec}}$ ?

(Report magnitudes; the electric force is attractive because charges are opposite.)

$F_{\text{grav}} \approx 1.3 \times 10^{-9}\,\text{N}$ and $F_{\text{elec}} \approx 0.90\,\text{N}$ (correct answer)
$F_{\text{grav}} \approx 1.3 \times 10^{-9}\,\text{N}$ and $F_{\text{elec}} \approx 900\,\text{N}$
$F_{\text{grav}} \approx 1.3 \times 10^{-1}\,\text{N}$ and $F_{\text{elec}} \approx 0.90\,\text{N}$
$F_{\text{grav}} \approx 1.3 \times 10^{-9}\,\text{N}$ and $F_{\text{elec}} \approx 9.0 \times 10^{-7}\,\text{N}$

2×10^{-6}

)(

2×10^{-6}

)/(0.2)² ≈ 0.9 N (noticeable), a ratio of about 10^9—even this modest laboratory charge produces electric forces that are a billion times stronger than gravity between the same objects. Choice A is correct because it properly applies both force formulas and compares the resulting magnitudes. Choice D has an error in the power of 10, reporting F_elec as 9.0 × 10^{-7} instead of 0.9, likely from incorrectly handling exponents when dividing powers of 10 in the calculation: (10⁹)(

10^{-12}

) / (

4×10^{-2}

) = 10^9 × 10^{-12} / 4×10^{-2} = (

10^{-3}

) / 4×10^{-2} wait, actually miscalculating the charge product. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). Practical implications: chemistry, molecular biology, material strength, friction, and essentially all everyday phenomena (except falling) are determined by electric forces between atoms and molecules, while planetary orbits, tides, satellite motion, and the large-scale structure of the universe are determined by gravitational forces—the electric force's dominance at small scales is why a charged balloon can lift paper against Earth's entire gravitational pull, yet gravity's dominance at large scales is why planets orbit stars despite any residual electric charges they might have.

Question 13

(Recall: $F_{\text{grav}} = G\dfrac{m_1m_2}{r^2}$ and $F_{\text{elec}} = k\dfrac{|q_1q_2|}{r^2}$ .)

$2.3 \times 10^{-39}$
$2.3 \times 10^{29}$
$2.3 \times 10^{39}$ (correct answer)
$2.3 \times 10^{49}$

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = kq₁q₂/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. Additionally, gravity is always attractive (masses always pull together) while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For a proton and electron separated by the Bohr radius r ≈ 5.3 × 10⁻¹¹ m, the gravitational force is F_grav = G(m_p)(m_e)/r² = (6.67×10⁻¹¹)(1.67×10⁻²⁷)(9.11×10⁻³¹)/(5.3×10⁻¹¹)² ≈ 3.6 × 10⁻⁴⁷ N, while the electric force is F_elec = k(e)(e)/r² = (9.0×10⁹)(1.6×10⁻¹⁹)²/(5.3×10⁻¹¹)² ≈ 8.2 × 10⁻⁸ N. The ratio F_elec/F_grav ≈ (8.2×10⁻⁸)/(3.6×10⁻⁴⁷) ≈ 2.3 × 10³⁹, meaning the electric force is about 10³⁹ times stronger—this enormous difference explains why gravity is completely negligible at atomic scales and electron orbits are determined entirely by electric attraction. Choice C is correct because it properly calculates the ratio (kq₁q₂)/(Gm₁m₂) with correct scientific notation. Choice A inverts the ratio, calculating F_grav/F_elec instead of F_elec/F_grav, which gives a tiny number like 10⁻³⁹ instead of the enormous ratio 10³⁹, incorrectly suggesting gravity is stronger when actually electric force is vastly stronger at this scale. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav = G(M_planet)(M_star)/r² enormous while F_elec ≈ k(~0)(~0)/r² ≈ 0.

Question 14

Two protons are separated by $r = 1.0 \times 10^{-15}\,\text{m}$ . Use $m_p = 1.67 \times 10^{-27}\,\text{kg}$ , $q_p = +1.6 \times 10^{-19}\,\text{C}$ , $G = 6.67 \times 10^{-11}$ , and $k = 9.0 \times 10^9$ . Which pair of magnitudes is closest to the gravitational attraction and electric repulsion between them?

Compute $F_{\text{grav}} = G\dfrac{m_p^2}{r^2}$ and $F_{\text{elec}} = k\dfrac{q_p^2}{r^2}$ .

$F_{\text{grav}} \approx 1.9 \times 10^{-34}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{2}\,\text{N}$ (correct answer)
$F_{\text{grav}} \approx 1.9 \times 10^{-14}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{-18}\,\text{N}$
$F_{\text{grav}} \approx 1.9 \times 10^{-34}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{-18}\,\text{N}$
$F_{\text{grav}} \approx 1.9 \times 10^{2}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{-34}\,\text{N}$

10^{-30}

) for r² gives large positive exponent. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav = G(M_planet)(M_star)/r² enormous while F_elec ≈ k(~0)(~0)/r² ≈ 0.

Question 15

Two objects are separated by the same distance $r$ . Object 1 has mass $m_1$ and charge $q_1$ , and object 2 has mass $m_2$ and charge $q_2$ . Which expression correctly gives the ratio of the magnitude of the electric force to the magnitude of the gravitational force between them?

Use $F_{\text{grav}} = G\dfrac{m_1m_2}{r^2}$ and $F_{\text{elec}} = k\dfrac{|q_1q_2|}{r^2}$ .

$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{k|q_1q_2|}{Gm_1m_2}$ (correct answer)
$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{k|q_1q_2|r^2}{Gm_1m_2}$
$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{Gm_1m_2}{k|q_1q_2|}$
$\dfrac{F_{\text{elec}}}{F_{\text{grav}}} = \dfrac{k|q_1q_2|}{Gm_1m_2r^2}$

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = kq₁q₂/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. Additionally, gravity is always attractive (masses always pull together) while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For scale dependence: The ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of distance r (since both forces have r² in denominator which cancels in the ratio), meaning the relative strength depends only on the charge-to-mass ratios of the objects—this is why electric forces dominate at atomic scales (where charge-to-mass ratio is high for particles like electrons) while gravity dominates at cosmic scales (where objects are electrically neutral with charge-to-mass ratio near zero, but masses are enormous). Choice A is correct because it properly applies both force formulas and compares the resulting magnitudes. Choice D incorrectly claims that the distance dependence is different for the two forces (claiming one decreases faster with distance than the other), when actually both follow inverse square laws (F ∝ 1/r²) and decrease at the same rate with distance—the ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of r. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). Practical implications: chemistry, molecular biology, material strength, friction, and essentially all everyday phenomena (except falling) are determined by electric forces between atoms and molecules, while planetary orbits, tides, satellite motion, and the large-scale structure of the universe are determined by gravitational forces—the electric force's dominance at small scales is why a charged balloon can lift paper against Earth's entire gravitational pull, yet gravity's dominance at large scales is why planets orbit stars despite any residual electric charges they might have.

Question 16

Two identical metal spheres each have mass $m = 1.0\,\text{kg}$ and net charge $q = +1.0\,\mu\text{C} = 1.0\times 10^{-6}\,\text{C}$ . Their centers are $r = 0.50\,\text{m}$ apart. Using $G = 6.67 \times 10^{-11}$ and $k = 9.0 \times 10^9$ , which statement best describes which force dominates and by approximately what factor (magnitude ratio)?

Gravity dominates by about $10^{8}$ because masses are much larger than charges.
Electric dominates by about $10^{11}$ because $F_{\text{elec}}/F_{\text{grav}} \approx \dfrac{kq^2}{Gm^2}$ . (correct answer)
They are equal because both follow an inverse-square law ( $1/r^2$ ).
Electric dominates by about $10^{-11}$ because $k \ll G$ .

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = kq₁q₂/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. Additionally, gravity is always attractive (masses always pull together) while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For macroscopic charged objects: For two 1 kg objects each with charge 1 μC = 10⁻⁶ C separated by 0.5 m, the gravitational force is F_grav = (6.67×10⁻¹¹)(1)(1)/(0.5)² ≈ 2.7 × 10⁻¹⁰ N (tiny), while the electric force is F_elec = (9.0×10⁹)(10⁻⁶)(10⁻⁶)/(0.5)² = 3.6 × 10^{-2} N (noticeable), a ratio of about 10^{11}—even this modest laboratory charge produces electric forces that are a hundred billion times stronger than gravity between the same objects. Choice B is correct because it accurately identifies that electric force dominates at atomic scale due to the enormous force ratio of ~10¹¹ and properly applies both force formulas and compares the resulting magnitudes. Choice D has an error in the power of 10, reporting the ratio as 10^{-11} instead of the correct 10^{11}, likely from incorrectly handling exponents when dividing powers of 10 in the calculation: (10⁹)/(10⁻¹¹) = 10²⁰ for the constants, combined with exponents from masses and charges. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). Practical implications: chemistry, molecular biology, material strength, friction, and essentially all everyday phenomena (except falling) are determined by electric forces between atoms and molecules, while planetary orbits, tides, satellite motion, and the large-scale structure of the universe are determined by gravitational forces—the electric force's dominance at small scales is why a charged balloon can lift paper against Earth's entire gravitational pull, yet gravity's dominance at large scales is why planets orbit stars despite any residual electric charges they might have.

Question 17

Two protons are separated by $r = 1.0 \times 10^{-15}\,\text{m}$ . Use $m_p = 1.67 \times 10^{-27}\,\text{kg}$ , $q_p = +1.6 \times 10^{-19}\,\text{C}$ , $G = 6.67 \times 10^{-11}$ , and $k = 9.0 \times 10^9$ . Which pair of magnitudes is closest to the gravitational attraction and electric repulsion between them?

Compute $F_{\text{grav}} = G\dfrac{m_p^2}{r^2}$ and $F_{\text{elec}} = k\dfrac{q_p^2}{r^2}$ .

$F_{\text{grav}} \approx 1.9 \times 10^{-34}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{2}\,\text{N}$ (correct answer)
$F_{\text{grav}} \approx 1.9 \times 10^{-14}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{-18}\,\text{N}$
$F_{\text{grav}} \approx 1.9 \times 10^{-34}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{-18}\,\text{N}$
$F_{\text{grav}} \approx 1.9 \times 10^{2}\,\text{N}$ and $F_{\text{elec}} \approx 2.3 \times 10^{-34}\,\text{N}$

10^{-30}

Question 18

(Report magnitudes; the electric force is attractive because charges are opposite.)

$F_{\text{grav}} \approx 1.3 \times 10^{-9}\,\text{N}$ and $F_{\text{elec}} \approx 0.90\,\text{N}$ (correct answer)
$F_{\text{grav}} \approx 1.3 \times 10^{-9}\,\text{N}$ and $F_{\text{elec}} \approx 900\,\text{N}$
$F_{\text{grav}} \approx 1.3 \times 10^{-1}\,\text{N}$ and $F_{\text{elec}} \approx 0.90\,\text{N}$
$F_{\text{grav}} \approx 1.3 \times 10^{-9}\,\text{N}$ and $F_{\text{elec}} \approx 9.0 \times 10^{-7}\,\text{N}$

2×10^{-6}

)(

2×10^{-6}

10^{-12}

) / (

4×10^{-2}

) = 10^9 × 10^{-12} / 4×10^{-2} = (

10^{-3}

Question 19

Consider the Earth and the Moon separated by about $r = 3.8 \times 10^8\,\text{m}$ . Suppose (hypothetically) both bodies have extremely small net charges compared with their total number of protons/electrons, so their net charges are effectively $q_{\oplus} \approx 0$ and $q_{\text{Moon}} \approx 0$ . Which statement best explains why gravity dominates their interaction even though the electric force constant $k$ is much larger than $G$ ?

Because electric forces only act at short range, while gravity acts at long range.
Because large astronomical bodies are nearly electrically neutral, so the net electric force is near zero while gravity adds from all the mass. (correct answer)
Because gravity can be either attractive or repulsive, canceling out electric effects.
Because the electric force decreases as $1/r^3$ but gravity decreases as $1/r^2$ .

Explanation: This question tests understanding of how gravitational and electric forces compare in magnitude and significance at different scales. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = kq₁q₂/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values. Additionally, gravity is always attractive (masses always pull together) while electric force can be attractive (opposite charges) or repulsive (same charges), and gravity depends on mass while electric force depends on charge. For scale dependence: The ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of distance r (since both forces have r² in denominator which cancels in the ratio), meaning the relative strength depends only on the charge-to-mass ratios of the objects—this is why electric forces dominate at atomic scales (where charge-to-mass ratio is high for particles like electrons) while gravity dominates at cosmic scales (where objects are electrically neutral with charge-to-mass ratio near zero, but masses are enormous). Choice B is correct because it correctly explains that gravity dominates at cosmic scale because massive objects are electrically neutral. Choice A suggests that the distance dependence is different for the two forces (claiming one decreases faster with distance than the other), when actually both follow inverse square laws (F ∝ 1/r²) and decrease at the same rate with distance—the ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of r. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). The key insight is that electric force is far stronger intrinsically (k >> G by 10²⁰), yet gravity controls the cosmos—this apparent paradox resolves when you realize that charge comes in two types (+ and -) that cancel when combined, while mass comes in only one type (positive) that always adds up, so large objects like planets inevitably have huge total mass but near-zero net charge, making F_grav = G(M_planet)(M_star)/r² enormous while F_elec ≈ k(~0)(~0)/r² ≈ 0.

Question 20

Two objects are separated by the same distance $r$ . Object 1 has $m_1 = 2.0\ \text{kg}$ and $q_1 = +3.0\ \mu\text{C}$ ; object 2 has $m_2 = 4.0\ \text{kg}$ and $q_2 = -3.0\ \mu\text{C}$ . Which expression correctly gives the ratio of magnitudes $\left|F_{\text{elec}}\right|/\left|F_{\text{grav}}\right|$ for this pair?

$\dfrac{k\,|q_1 q_2|}{G\,m_1 m_2}$ (correct answer)
$\dfrac{G\,m_1 m_2}{k\,|q_1 q_2|}$
$\dfrac{k\,|q_1 q_2|\,r^2}{G\,m_1 m_2}$
$\dfrac{k\,|q_1 q_2|}{G\,m_1 m_2\,r^2}$