Compare Gravitational and Electric Forces - Grade 11 Physics

1. How much work is done by the combined electric and gravitational forces during this motion?
   Answer: The combined forces do work equal to $+2U_0$. When you see a question involving both gravitational and electric forces, the key framework is the work-energy theorem: the work done by conservative forces equals the negative change in potential energy, or equivalently, $W = U_i - U_f$. Start by establishing the initial total potential energy. You're told the gravitational PE has magnitude $U_0$ and the electric PE is $4U_0$. Since gravity between like charges is attractive (negative PE) and the electric force is repulsive (positive PE), the initial total is $U_i = 4U_0 - U_0 = 3U_0$. Now find the final energies at separation $3r$. Both gravitational and electric potential energies scale as $1/r$, so tripling the distance divides each by 3. The electric PE becomes $\frac{4U_0}{3}$ and the gravitational PE magnitude becomes $\frac{U_0}{3}$, giving $U_f = \frac{4U_0}{3} - \frac{U_0}{3} = \frac{3U_0}{3} = U_0$. The work done is $W = U_i - U_f = 3U_0 - U_0 = +2U_0$, confirming C is correct. As for the distractors: A ($+3U_0$) ignores gravity entirely and also doesn't apply the $1/r$ scaling correctly. B ($-2U_0$) likely reverses the formula, computing $U_f - U_i$ instead of $U_i - U_f$. D ($-3U_0$) compounds errors — perhaps using only initial values with the wrong sign convention. Study tip: Always track signs carefully — gravitational PE between masses is negative, while electric PE between like charges is positive. Writing out $W = U_i - U_f$ explicitly before plugging in numbers prevents the sign-flip mistake that traps many students.
2. Equal-magnitude $F_e$ and $F_g$ act on two charges. If $q_1q_2=3\times10^{-16}$ C$^2$, then $m_1m_2$ (kg$^2$) is about
   Answer: $4.1\times10^{4}$ Set k q1 q2 / $r^2$ = G m1 m2 / $r^2$. The $r^2$ cancels, so m1 m2 = (k/G) q1 q2 = (9e9 / 6.67e-11)(3e-16), which is about 4.1e4 $kg^2$. The tempting 2.2e-36 comes from using G/k instead of k/G.
3. What electric field is required for the electric force to balance the gravitational force?
   Answer: A downward field of magnitude $1.96\times10^4\ \mathrm{N/C}$. When a charged particle is in equilibrium, the net force on it must be zero — meaning the electric force must exactly cancel gravity. The gravitational force pulls the drop downward with magnitude $F_g = mg = (4.0\times10^{-15})(9.8) = 3.92\times10^{-14}\ \mathrm{N}$. So the electric force must point upward with the same magnitude. Here's the critical twist: the drop carries a negative charge. The electric force on a charge is $\vec{F} = q\vec{E}$. For a negative charge, the force is opposite to the field direction. So to produce an upward electric force on a negative charge, the field itself must point downward. The magnitude of the required field is: $E = \frac{F_g}{|q|} = \frac{3.92\times10^{-14}}{2.0\times10^{-18}} = 1.96\times10^4\ \mathrm{N/C}$ This confirms D — a downward field of $1.96\times10^4\ \mathrm{N/C}$. A gets the direction wrong (upward field would push the negative drop downward, worsening the imbalance) and also uses the wrong magnitude. B has the correct magnitude but the wrong direction — a very common trap for students who find $E$ correctly but forget that $\vec{F} = q\vec{E}$ reverses direction for negative charges. C has the right direction but the wrong magnitude, likely from a computational error such as using incorrect values of mass or charge. Your study tip: always apply $\vec{F} = q\vec{E}$ with the sign of the charge before deciding field direction. For negative charges, force and field point in opposite directions — this single step trips up many students.
4. At what separation will the electric and gravitational forces have equal magnitudes?
   Answer: At no separation, because both force magnitudes have the same inverse-square dependence. Whenever you see a question comparing two forces that both follow the inverse-square law, the key insight is recognizing how their ratio behaves as distance changes. Both the electric force and gravitational force follow the same mathematical form: $F_E = k\frac{q_1 q_2}{r^2}, \quad F_G = G\frac{m_1 m_2}{r^2}$ Notice that both forces depend on $r^2$ in exactly the same way. When you form the ratio of these two forces, the $r^2$ terms cancel completely: $\frac{F_E}{F_G} = \frac{kq_1q_2}{Gm_1m_2} = \text{constant}$ This ratio depends only on the particles' charges and masses — which the problem tells you remain fixed. If the electric force is five times the gravitational force at $0.20\ \mathrm{m}$, it will be five times as large at any separation. The two forces scale identically with distance, so their ratio never changes. They can never be equal unless they were equal to begin with. This makes D the correct answer. A is wrong because it incorrectly suggests the two forces have different distance dependences. They don't — both go as $1/r^2$. B makes a similar error, assuming you can "tune" the ratio by changing separation, but since both forces shrink or grow at the same rate, changing $r$ accomplishes nothing. C is wrong for the same fundamental reason — multiplying separation by any factor affects both forces identically, leaving their ratio unchanged. The strategy here: when two forces share the same $r$-dependence, their ratio is distance-independent. Any question asking "at what distance do they become equal?" has the answer "never" — unless they start equal.
5. How does the electric force between the grains compare with their gravitational force?
   Answer: The electric force is repulsive and approximately $3.0\times10^{10}$ times as large. When a question asks you to compare two fundamental forces, your job is to calculate each force using the same separation distance and then find their ratio — conveniently, the distance cancels out completely. Both forces follow an inverse-square law: the gravitational force is $F_g = G\frac{m_1 m_2}{r^2}$ and the electric force is $F_e = k\frac{q_1 q_2}{r^2}$. Taking the ratio eliminates $r^2$: $\frac{F_e}{F_g} = \frac{k\, q^2}{G\, m^2} = \frac{(8.99\times10^9)(3.0\times10^{-14})^2}{(6.67\times10^{-11})(2.0\times10^{-9})^2}$ Numerator: $(8.99\times10^9)(9.0\times10^{-28}) \approx 8.09\times10^{-18}\ \mathrm{N\,m^2}$ Denominator: $(6.67\times10^{-11})(4.0\times10^{-18}) \approx 2.67\times10^{-28}\ \mathrm{N\,m^2}$ $\frac{F_e}{F_g} \approx \frac{8.09\times10^{-18}}{2.67\times10^{-28}} \approx 3.0\times10^{10}$ Since both grains carry positive charge, the electric force is repulsive, confirming answer A. Answer B is wrong because like charges repel — the electric force cannot be attractive here. Answer C incorrectly identifies gravity (not the electric force) as the dominant force and inverts the comparison entirely. Answer D flips the ratio upside down, mistakenly suggesting the electric force is the weaker one by a factor of $3.3\times10^{-11}$. A useful study tip: whenever you compare two forces, set up the ratio immediately — shared variables like distance cancel, simplifying the arithmetic significantly. Also remember that "like charges repel" is a fundamental rule you should apply before doing any calculation.
6. Which statement best describes the electric and gravitational forces between the particles?
   Answer: Both forces are attractive, and the electric force is approximately $2.3\times10^{39}$ times stronger. When comparing fundamental forces between subatomic particles, you need to evaluate both the direction and relative magnitude of each force separately before drawing conclusions. For direction: gravity is always attractive between any two masses, and the electric force between opposite charges (proton positive, electron negative) is also attractive. This immediately eliminates C, which incorrectly claims electricity is repulsive between a proton and electron — opposite charges always attract. For magnitude, both forces depend on the same separation distance $r$, so that cancels when you take the ratio. The ratio of electric to gravitational force is: $\frac{F_e}{F_g} = \frac{ke^2}{Gm_pm_e} = \frac{(8.99\times10^9)(1.60\times10^{-19})^2}{(6.67\times10^{-11})(1.67\times10^{-27})(9.11\times10^{-31})}$ Numerator: $8.99\times10^9 \times 2.56\times10^{-38} \approx 2.30\times10^{-28}$ Denominator: $6.67\times10^{-11} \times 1.52\times10^{-57} \approx 1.01\times10^{-67}$ $\frac{F_e}{F_g} \approx 2.3\times10^{39}$ This confirms A — both forces are attractive, and the electric force is roughly $2.3\times10^{39}$ times stronger than gravity. B has the ratio inverted, claiming gravity is stronger — a serious misconception, since gravity between subatomic particles is negligibly weak. D uses the reciprocal value ($4.4\times10^{-40}$), which would mean electricity is weaker than gravity — the opposite of reality. Study tip: When comparing forces, always compute the ratio $F_e/F_g$ — the distance cancels cleanly, making this a straightforward plug-and-chug. Remember: at the atomic scale, electric force utterly dominates gravity.
7. For two charged particles, $F_e=9.2\times10^{-8}$ N and $F_e/F_g=2.3\times10^{39}$. Then $F_g$ is
   Answer: $4.0\times10^{-47}$ N Since Fe/Fg = 2.3 x $10^{39}$, divide Fe by that ratio: (9.2 x $10^{-8}$) / (2.3 x $10^39$) = 4.0 x $10^{-47}$ N. The tempting wrong move is multiplying instead of dividing, which gives 2.1 x $10^{32}$ N, but the ratio tells you Fe is much larger than Fg, so Fg must be tiny.
8. At the new altitude, what is the net force on the probe expressed in terms of the local gravitational force magnitude $F_g$?
   Answer: $1.4F_g$ upward, because the unchanged electric force now exceeds the reduced local gravity. When a problem involves multiple forces at different altitudes, your job is to track each force independently — don't assume they all scale together just because altitude changes. At Earth's surface, let the probe's weight be $W$. The electric force is $F_E = 0.60W$ upward. At the new altitude, gravity drops to $0.25$ of its surface value, so the gravitational force becomes $F_g = 0.25W$. Crucially, the problem states the electric field and charge are unchanged, so the electric force remains $0.60W$ — it doesn't scale with altitude the way gravity does. Now express the electric force in terms of the local gravitational force $F_g = 0.25W$: since $0.60W = 2.4 \times (0.25W)$, the electric force equals $2.4F_g$ upward. The net force is then $2.4F_g - F_g = 1.4F_g$ upward. That confirms B is correct. Choice A is wrong because it assumes the electric force shrinks with gravity — but electric forces depend on the field and charge, not on altitude-related gravitational changes. Choice C confuses the electric-to-local-gravity ratio ($2.4F_g$) with the net force, forgetting to subtract gravity itself. Choice D incorrectly concludes that gravity still dominates; at this altitude, the electric force (still $0.60W$) far exceeds the weakened gravitational force (only $0.25W$). A useful habit: always list every force and ask "does this force actually change with the new condition?" Electric forces are set by field strength and charge — not by gravitational field strength.
9. After these changes, how does the electric force compare with the gravitational force?
   Answer: The electric force remains attractive and has magnitude $6F_g$. When a question changes multiple variables at once, your job is to track each force separately using its own formula before comparing them. For the electric force, recall Coulomb's Law: $F_e = k\frac{q_1 q_2}{r^2}$. The charge $q_1$ triples (factor of 3) and the separation halves (factor of $\frac{1}{2}$, which squares to $\frac{1}{4}$ in the denominator, giving a factor of 4). So the electric force scales by $3 \times 4 = 12$. The new electric force is $12 \times$ the original electric force. For the gravitational force, recall Newton's Law: $F_g = G\frac{m_1 m_2}{r^2}$. The mass $m_1$ doubles (factor of 2) and the separation halves (same factor of 4 as above). So the gravitational force scales by $2 \times 4 = 8$. The new gravitational force is $8F_g$. Now compare: the new electric force is $12 \times$ the original electric force. The original electric force was $4F_g$, so the new electric force equals $4F_g \times 12 = 48$ in absolute terms. Divide by the new gravitational force ($8F_g$): $\frac{48}{8} = 6$. The electric force is $6$ times the new gravitational force, confirming answer B. Answer A incorrectly ignores all the changes. Answer D correctly finds the new electric force scales by 12 but forgets to account for the gravitational force also changing — it mistakenly compares the new electric force to the original $F_g$. Answer C introduces a sign error; tripling a charge on one object doesn't flip the sign of the other object's charge, so attraction remains. Always ask yourself: "Did both forces change?" — if the separation changes, it affects electric and gravitational forces equally, but mass and charge changes are independent.
10. Two identical spheres have $F_e=F_g$. If each charge doubles and $r$ halves, what new mass restores equality?
    Answer: $2m$ Starting from Fe=Fg, doubling both charges and halving separation makes electric force 16 times larger (2 squared divided by 1/2 squared). The halved separation alone also makes gravitational force 4 times larger, so the mass-squared term must add another factor of 4. Since gravity depends on mass squared, each mass becomes 2m. The trap answer 4m ignores that gravity already gained a factor of 4 from the shorter distance.
11. For two electrons, $F_e/F_g=R$. What separation between two protons gives the same $R$?
    Answer: No separation can work The electric and gravitational forces both obey the inverse-square law, so distance cancels in Fe/Fg. For electrons, R = k $e^2$ / (G m_$e^2$). For protons, the same expression becomes k $e^2$ / (G m_$p^2$) = $R(m_e/m_p)^2$. Since separation never changes the ratio, no distance between protons can make it equal R. The tempting idea that the same separation works ignores that proton mass differs.
12. Approximately what charge must be placed on each sphere?
    Answer: Each sphere must carry approximately $1.7\times10^{-11}\ \mathrm{C}$. When two forces are asked to "exactly balance," your job is to set them equal and solve. Here, you're balancing the gravitational attraction between two masses against the electrostatic repulsion between two charges. Both forces follow an inverse-square law, which means the separation distance cancels out — a helpful simplification. Set the two force equations equal: $F_g = F_e \implies G\frac{m^2}{r^2} = k\frac{q^2}{r^2}$ Since $r^2$ appears in both denominators, it cancels, leaving: $Gm^2 = kq^2 \implies q^2 = \frac{G}{k}m^2 \implies q = m\sqrt{\frac{G}{k}}$ Plugging in values: $q = (0.20)\sqrt{\frac{6.67\times10^{-11}}{8.99\times10^{9}}} = (0.20)\sqrt{7.42\times10^{-21}}$ $\sqrt{7.42\times10^{-21}} \approx 8.61\times10^{-11}$ $q \approx (0.20)(8.61\times10^{-11}) \approx 1.7\times10^{-11}\ \mathrm{C}$ This confirms C is correct. Choice A ($3.0\times10^{-22}\ \mathrm{C}$) likely results from forgetting to take the square root — leaving $q = m \cdot \frac{G}{k}$ instead of $q = m\sqrt{\frac{G}{k}}$. Choice B ($3.9\times10^{-11}\ \mathrm{C}$) comes from using $m = 0.20\ \mathrm{kg}$ as $m^2$ without squaring it properly, or a rounding error mid-calculation. Choice D ($1.7\times10^{-6}\ \mathrm{C}$) is off by five orders of magnitude, likely from a powers-of-ten arithmetic mistake when handling the small ratio under the square root. A reliable strategy: whenever distance cancels in a force-balance problem, immediately simplify before substituting numbers — it reduces arithmetic errors dramatically.
13. What speed must the charged satellite have to maintain a circular orbit at that radius?
    Answer: The charged satellite must travel at $0.80v_0$ because orbital speed depends on the square root of force. When a satellite orbits in a circle, the net inward (centripetal) force must equal $\frac{mv^2}{r}$. For a neutral satellite at speed $v_0$, gravity alone provides that force: $F_g = \frac{mv_0^2}{r}$. For the charged satellite, the outward electric force partially cancels gravity, so the net inward force becomes $F_g - 0.36F_g = 0.64F_g$. Setting this equal to the centripetal requirement: $0.64F_g = \frac{mv^2}{r}$. Since $F_g = \frac{mv_0^2}{r}$, you can substitute and divide: $\frac{v^2}{v_0^2} = 0.64$, which gives $v = \sqrt{0.64}\, v_0 = 0.80\, v_0$. So the charged satellite must travel at **$0.80v_0$** — answer B. Choice A reaches the right net force ($0.64F_g$) but then forgets to take the square root, incorrectly claiming $v = 0.64v_0$. Orbital speed depends on the square root of the net force, not the force directly. Choice C treats the electric force as adding to gravity rather than opposing it. Since both the satellite and planet are positively charged, the electric force is repulsive — it points outward, reducing the net centripetal force. Choice D makes the same direction error as C and also skips the square root relationship, compounding two mistakes. The key strategy here: always identify the direction of each force first (inward = helps orbit, outward = hurts it), find the net centripetal force, then remember that $v \propto \sqrt{F_{\text{net}}}$ — never a direct proportion.
14. For an electron-proton pair, the ratio $F_e/F_g$ is closest to
    Answer: $10^{39}$ Use Coulomb's law and Newton's law: Fe = k $e^2$ / $r^2$ and Fg = G me mp / $r^2$, so the $r^2$ cancels and Fe/Fg = k $e^2$ / (G me mp). Plugging in the values gives about 2.3 x $10^{39}$. The tempting trap is $10^{-39}$, which is the inverse ratio Fg/Fe; gravity is nowhere near that strong for an electron-proton pair.
15. Which statement correctly compares the net force before and after the charge reversal?
    Answer: The initial net force is $\frac{7}{3}$ times the final net force; initially it is attractive, and finally it is repulsive. When two forces act between objects — gravitational and electric — the net force is their vector sum, and you must account for both direction and magnitude carefully. Here's the setup: gravity is always attractive, pulling the droplets together. Before the reversal, the charges are opposite, so the electric force is also attractive. Both forces point the same direction, so they add: $F_{\text{before}} = F_g + F_e = F_g + 2.5F_g = 3.5F_g$ This net force is attractive. After reversing one charge's sign, both droplets now carry the same sign, making the electric force repulsive — it opposes gravity. Now the forces partially cancel: $F_{\text{after}} = F_e - F_g = 2.5F_g - F_g = 1.5F_g$ This net force is repulsive (electric wins). Taking the ratio: $\frac{F_{\text{before}}}{F_{\text{after}}} = \frac{3.5F_g}{1.5F_g} = \frac{7}{3}$ So the initial force is $\frac{7}{3}$ times the final force, initially attractive and finally repulsive — confirming answer A. Answer B is wrong because after reversal the two same-sign charges create a repulsive electric force that dominates gravity, so the net force is not attractive. Answer C gets the direction change right but miscalculates the ratio, likely by forgetting to include gravity in the "before" total. Answer D flips both the ratio and the directions entirely, a sign-confusion error. Study tip: Always identify the direction of each force before adding magnitudes. When forces oppose each other, subtract — when they align, add. Drawing a quick arrow diagram prevents the most common mistakes on combined-force problems.

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