Apply Coulomb's Law - Grade 11 Physics

1. Identical spheres $+5Q$ and $-3Q$ attract with $F$. Touch them; separate to original distance. New force?
   Answer: $F/15$, repulsive Initial pull comes from 15 $Q^2$ / $r^2$. After touching, net charge +5Q - 3Q = +2Q splits equally, so each sphere has +Q. New push is $Q^2$ / $r^2$ = F/15, and like charges repel. The tempting error is F/15 attractive: correct magnitude but both spheres end up positive, so the force cannot be attractive.
2. Two initially neutral, identical small spheres are separated by $0.40 m$. A total of $2.0 × 10^{12}$ electrons is transferred from sphere X to sphere Y. Treating the spheres as point charges, what force acts between them? Use an electron-charge magnitude of $1.6 × 10^{-19} C$ and $k = 9.0 × 10^9 N·m^2/C^2$.
   Answer: $5.8 × 10^{-3} N$, attractive When you see a question involving electron transfer between neutral spheres, your two tasks are: (1) find the charge on each sphere, then (2) apply Coulomb's Law — and don't forget to determine whether the force is attractive or repulsive. Finding the charge: Each electron carries $1.6 \times 10^{-19} \text{ C}$, so transferring $2.0 \times 10^{12}$ electrons gives a total charge of: $q = (2.0 \times 10^{12})(1.6 \times 10^{-19}) = 3.2 \times 10^{-7} \text{ C}$ Sphere Y gains electrons, making it negative ($-3.2 \times 10^{-7} \text{ C}$), while sphere X loses them, becoming positive ($+3.2 \times 10^{-7} \text{ C}$). Opposite charges → attractive force. Applying Coulomb's Law: $F = k\frac{q_1 q_2}{r^2} = (9.0 \times 10^9)\frac{(3.2 \times 10^{-7})^2}{(0.40)^2}$ $F = (9.0 \times 10^9)\frac{1.024 \times 10^{-13}}{0.16} \approx 5.8 \times 10^{-3} \text{ N}$ This confirms answer A — $5.8 \times 10^{-3} \text{ N}$, attractive. B gets the magnitude right but says repulsive — a critical error. Since one sphere is positive and the other negative, the force must be attractive, not repulsive. C likely results from forgetting to square the charge (or using the wrong value of $r$), giving an incorrectly small force. D may come from arithmetic errors, such as using $r = 0.40$ instead of $r^2 = 0.16$. Study tip: Always label each sphere's sign before plugging into Coulomb's Law — the direction (attractive vs. repulsive) is just as important as the magnitude and is a frequent trap on this exam.
3. Two point charges exert an electrostatic force of magnitude $F$ on each other. The first charge is doubled, the second charge is reduced to half its original value, and the separation between the charges is tripled. What is the magnitude of the new electrostatic force?
   Answer: $F/9$ Whenever you see a question about changing the variables in Coulomb's Law, your first move should be to write out the ratio of the new force to the original force rather than calculating absolute values. Coulomb's Law states $F = k\frac{q_1 q_2}{r^2}$. When the first charge doubles, the second charge becomes half, and the separation triples, the new force is: $F' = k\frac{(2q_1)(\frac{1}{2}q_2)}{(3r)^2}$ Simplify step by step. The charges multiply together: $(2)(\frac{1}{2}) = 1$, so the numerator is unchanged — it's still $kq_1q_2$. The denominator becomes $(3r)^2 = 9r^2$. Therefore: $F' = \frac{kq_1q_2}{9r^2} = \frac{F}{9}$ The correct answer is A. Looking at the wrong choices: B ($F/3$) reflects a common error where students forget to square the new separation — they divide by 3 instead of 9. C ($F$) comes from correctly canceling the charge changes (which do multiply to 1) but then ignoring the effect of tripling the distance entirely. D ($3F$) suggests the student inverted the distance effect, treating the force as proportional to distance rather than inversely proportional to distance squared. A reliable strategy: handle each variable change one at a time as a multiplying factor. Charges affect the numerator linearly; separation affects the denominator as a square. Writing $F' = F \times (\text{charge factor}) \times \frac{1}{(\text{distance factor})^2}$ keeps you from mixing up which operations to apply.
4. Two fixed positive point charges, $+4.0 × 10^{-6} C$ and $+1.0 × 10^{-6} C$, are separated by $0.90 m$. A small positive test charge is placed on the line between them. At what distance from the $+4.0 × 10^{-6} C$ charge will the net electrostatic force on the test charge be zero?
   Answer: $0.60 m$ When two like charges repel a test charge from both sides, there's exactly one point between them where those forces cancel out — and it's always closer to the weaker charge. This is the key insight for equilibrium-of-forces problems. Let the distance from the $+4.0 \times 10^{-6}\text{ C}$ charge be $d$, so the distance from the $+1.0 \times 10^{-6}\text{ C}$ charge is $(0.90 - d)$. Setting the two Coulomb forces on the test charge equal: $\frac{kq_1}{d^2} = \frac{kq_2}{(0.90-d)^2}$ The $k$ and test charge cancel, leaving: $\frac{4.0}{d^2} = \frac{1.0}{(0.90-d)^2}$ Taking the square root of both sides: $\frac{2}{d} = \frac{1}{0.90-d}$, which gives $2(0.90 - d) = d$, so $1.80 = 3d$, and $d = 0.60\text{ m}$. This confirms C is correct. Choice A (0.30 m) reflects a common error of dividing the total distance in ratio 4:1 rather than 2:1 — forgetting to take the square root of the charge ratio. Choice B (0.45 m) is simply the midpoint, which would only be correct if both charges were equal. Choice D (0.75 m) places the point closer to the stronger charge, which is physically backwards — the stronger charge's force reaches equilibrium farther away, not closer. As a strategy tip: always take the square root of the charge ratio when solving for the equilibrium position, and double-check that your answer sits closer to the weaker charge.
5. A positive charge $q$ is midway between $+Q$ and $-Q$ (separation $d$). Net force on $q$ is
   Answer: $8kQq/d^2$ toward $-Q$ At the midpoint, q is distance d/2 from each charge. The +Q repels q away from it, while the -Q attracts q toward it, so both forces point toward -Q. Each force has magnitude kQq/$(d/2)^2$ = 4kQq/$d^2$, so they add to 8kQq/$d^2$. The tempting wrong answer is zero, but the forces are in the same direction, not opposite.
6. Using $g = 9.8 m/s^2$ and $k = 9.0 × 10^9 N·m^2/C^2$, what is the charge on each sphere?
   Answer: $9.8 × 10^{-8} C$ When you see charged spheres hanging from strings, you're dealing with electrostatic equilibrium — the electric repulsion force balances with gravity through the string tension. The key is analyzing the forces on one sphere using trigonometry, then applying Coulomb's Law. For each sphere in equilibrium, three forces act: tension $T$ (along the string), gravity $mg$ (downward), and electric force $F_E$ (horizontal). From the geometry: $\tan\theta = \frac{F_E}{mg}$. Plugging in the values: $F_E = mg\tan(5.0°) = (0.010)(9.8)\tan(5.0°) = 0.098 × 0.0875 \approx 8.57 × 10^{-3} \text{ N}$ Now apply Coulomb's Law, where both spheres carry equal charge $q$: $F_E = \frac{kq^2}{r^2} \Rightarrow q = \sqrt{\frac{F_E \cdot r^2}{k}} = \sqrt{\frac{(8.57×10^{-3})(0.10)^2}{9.0×10^9}}$ $q = \sqrt{\frac{8.57×10^{-5}}{9.0×10^9}} = \sqrt{9.52×10^{-15}} \approx 9.8×10^{-8} \text{ C}$ This confirms answer B. Choice A ($3.1×10^{-8}$ C) results from using $\sin\theta$ instead of $\tan\theta$, a common trigonometry mistake. Choice C ($3.1×10^{-7}$ C) likely comes from forgetting to square the separation distance in Coulomb's Law. Choice D ($2.8×10^{-7}$ C) may result from using the full separation as radius without squaring it correctly in the square root step. Study tip: Always draw a free-body diagram first. The ratio $\tan\theta = F_E/mg$ is your bridge between geometry and forces — memorize this setup for any suspended-charge problem.
7. Using $k = 9.0 × 10^9 N·m^2/C^2$, what is the net electrostatic force on the middle charge?
   Answer: $1.25 N$ to the right When dealing with net electrostatic force on a charge, your job is to calculate each individual force separately using Coulomb's Law, assign directions, then add them as vectors (paying close attention to signs and directions). Let's set up coordinates with right as positive. The middle charge ($+2.0 × 10^{-6}$ C) experiences forces from both neighboring charges. Force from the left charge ($+4.0 × 10^{-6}$ C, distance = 0.30 m): Both charges are positive, so they repel — this force pushes the middle charge to the right. $F_L = \frac{(9.0×10^9)(4.0×10^{-6})(2.0×10^{-6})}{(0.30)^2} = \frac{0.072}{0.09} = 0.80 \text{ N (right)}$ Force from the right charge ($-1.0 × 10^{-6}$ C, distance = 0.20 m): Opposite signs attract — this force also pulls the middle charge to the right. $F_R = \frac{(9.0×10^9)(2.0×10^{-6})(1.0×10^{-6})}{(0.20)^2} = \frac{0.018}{0.04} = 0.45 \text{ N (right)}$ Net force: $0.80 + 0.45 = 1.25 \text{ N to the right}$, confirming C is correct. Choice D gets the magnitude right but flips the direction — a sign error. Choices A and B (0.35 N) come from incorrectly subtracting the two forces instead of adding them, likely from assuming both forces act in opposite directions without checking the attraction/repulsion carefully. Your key strategy: always determine the direction of each force first based on charge signs before calculating magnitude. Sketch the setup — it prevents direction mistakes that trip up even careful students.
8. $+Q$ and $-4Q$ are separated by $L$ ($+Q$ on left). A positive charge feels zero net force at
   Answer: $L$ from $+Q$, to the left Set the distances so magnitudes cancel: Q/$d^2$ = 4Q/$(d+L)^2$. This gives $(d+L)^2$ = $4d^2$, so d+L = 2d and d = L. Thus the zero-force point is L to the left of +Q. The tempting wrong spot is between the charges, but there the +Q repels right and the -4Q attracts right, so the forces add instead of cancel.
9. Using $k = 9.0 × 10^9 N·m^2/C^2$, what is the net electrostatic force on the charge at the reference point?
   Answer: $0.75 N$, $36.9°$ north of west When solving net electrostatic force problems, your strategy is always the same: calculate each individual force separately, assign directions using vector components, then combine them using the Pythagorean theorem and trigonometry. Force from the +3.0 μC charge (east): Both charges are positive, so they repel. The reference charge gets pushed west. $F_1 = k\frac{q_1 q_2}{r^2} = \frac{(9.0×10^9)(2.0×10^{-6})(3.0×10^{-6})}{(0.30)^2} = \frac{0.054}{0.09} = 0.60 \text{ N, west}$ Force from the −4.0 μC charge (north): Opposite charges attract, so the reference charge gets pulled north. $F_2 = \frac{(9.0×10^9)(2.0×10^{-6})(4.0×10^{-6})}{(0.40)^2} = \frac{0.072}{0.16} = 0.45 \text{ N, north}$ Net force magnitude: $F_{net} = \sqrt{(0.60)^2 + (0.45)^2} = \sqrt{0.36 + 0.2025} = \sqrt{0.5625} = 0.75 \text{ N}$ Direction: The force points west and north, so the angle above the west axis is: $\theta = \tan^{-1}\left(\frac{0.45}{0.60}\right) = \tan^{-1}(0.75) = 36.9°\text{ north of west}$ This confirms answer A. Answer B is wrong because it reverses the direction — repulsion pushes away from the +3.0 μC charge (west), not east. Answer C incorrectly calculates the magnitude, likely by adding forces linearly instead of using the Pythagorean theorem. Answer D uses 53.1°, which is the complementary angle — a classic mix-up when students confuse which leg of the triangle is "opposite" versus "adjacent." Study tip: Always draw a vector diagram and label directions before calculating. Knowing why each force points a certain direction (attraction vs. repulsion) prevents the directional errors that eliminate answers B and D.
10. A bead of mass $2.0 × 10^{-3} kg$ is held at rest directly below a fixed charge of $+4.0 × 10^{-6} C$. The center-to-center separation is $0.20 m$, and no forces other than gravity and the electrostatic force act on the bead. Using $g = 9.8 m/s^2$ and $k = 9.0 × 10^9 N·m^2/C^2$, what charge must the bead have?
    Answer: $-2.2 × 10^{-8} C$ When a charged bead hangs in equilibrium, the net force on it is zero — meaning the upward electrostatic force must exactly balance the downward gravitational force. This is a classic static equilibrium problem combining Newton's second law with Coulomb's law. Start by calculating the gravitational force pulling the bead down: $F_g = mg = (2.0 \times 10^{-3})(9.8) = 1.96 \times 10^{-2} \text{ N}$ For the bead to be in equilibrium, the electrostatic force must equal this value and point upward. Since the fixed charge is positive and sits above the bead, it will repel a positive bead downward but attract a negative bead upward. So the bead must be negatively charged. Now solve for the bead's charge using Coulomb's law: $F_e = \frac{kq_1 q_2}{r^2} \Rightarrow q_2 = \frac{F_e \cdot r^2}{k \cdot q_1} = \frac{(1.96 \times 10^{-2})(0.20)^2}{(9.0 \times 10^9)(4.0 \times 10^{-6})}$ $q_2 = \frac{7.84 \times 10^{-4}}{3.6 \times 10^4} \approx 2.2 \times 10^{-8} \text{ C}$ Since it must be negative, the answer is **D: $-2.2 \times 10^{-8} \text{ C}$**. Choice A gives the correct magnitude but the wrong sign — the bead can't be positive, or it would be repelled downward, adding to gravity. Choice B is off by a factor of 5, likely from a unit or arithmetic error. Choice C comes from an incorrect intermediate calculation, possibly using the wrong separation distance. A reliable strategy: always determine the direction of the required force first, then use it to identify the sign of the charge before doing any arithmetic.
11. $+2q$ and $-3q$ attract with force $F$. Change $-3q$ to $+6q$ and double separation.
    Answer: $F/2$, repulsive The force scales with the product of the charges and inversely with separation squared. Initially the charge product is $6q^2$; after changing -3q to +6q it is $12q^2$, twice as large. Doubling the separation divides the force by 4, so the new force is 2/4 = 1/2 of F. Both charges are positive, so the force is repulsive. The tempting wrong answer is F/4, which ignores that the charge product doubled.
12. Original separation is $r$. One charge is quadrupled, the other tripled. To keep force constant, new separation is
    Answer: $2\sqrt{3}$ times original $r$ The charge product becomes 4 times 3 = 12 times larger, so the denominator $r^2$ must also become 12 times larger to keep force constant. That means new separation squared = 12 $r^2$, so new separation = sqrt(12) r = 2 sqrt(3) r. The tempting 12r is wrong because force changes with the square of distance, not linearly.
13. What must the unknown charge be so that the net electrostatic force on the charge at the origin is zero?
    Answer: $+4.50 × 10^{-6} C$ When a problem asks for the net electrostatic force to be zero, your goal is to find conditions where forces from all surrounding charges cancel each other out on the test charge. Here, the charge at the origin ($-1.0 \times 10^{-6}$ C) feels a force from the $+2.0 \times 10^{-6}$ C charge on the left and the unknown charge on the right — these two forces must be equal in magnitude and opposite in direction. First, calculate the force from the known charge (left side), using Coulomb's Law $F = k\frac{|q_1||q_2|}{r^2}$: $F_{left} = (9\times10^9)\frac{(2.0\times10^{-6})(1.0\times10^{-6})}{(0.20)^2} = 0.45 \text{ N (directed leftward, since opposite charges attract)}$ For the net force to be zero, the unknown charge must push or pull the origin charge rightward with equal magnitude. Since the origin charge is negative, a positive unknown charge would attract it rightward — perfect. Setting up: $0.45 = (9\times10^9)\frac{Q(1.0\times10^{-6})}{(0.30)^2}$ Solving gives $Q = +4.50 \times 10^{-6}$ C, confirming answer C. Answer A uses the correct sign but the wrong magnitude — likely from skipping the distance-squared factor or using wrong distances. Answer B makes the same magnitude error as A and also gets the sign wrong. Answer D gets the magnitude right but uses a negative sign, which would attract the origin charge leftward, adding to rather than canceling the existing force. Remember: always draw a force diagram first. Identify the direction each force acts before solving for magnitude — the sign of the unknown charge is determined by direction, not by formula alone.
14. Particle A has charge $+2.0 × 10^{-6} C$ and mass $m$. Particle B has charge $-6.0 × 10^{-6} C$ and mass $3m$. The particles are released from rest while separated by a fixed initial distance. Which statement correctly compares their initial electrostatic forces and accelerations?
    Answer: Their force magnitudes are equal, but A's acceleration is three times larger. Whenever you see a question pairing electrostatic force with acceleration, you need to apply two separate laws: Newton's Third Law and Newton's Second Law. Keep them distinct in your mind. Newton's Third Law tells you that the electrostatic force between two particles is an interaction pair — particle A pulls on B with exactly the same magnitude that B pulls on A, just in opposite directions. This is true regardless of charge magnitude or mass. So both particles experience the same force magnitude, which you can confirm with Coulomb's Law: $F = k\frac{|q_A||q_B|}{r^2}$. The charges and distance are identical for both calculations, giving one single force value. Now apply Newton's Second Law: $a = \frac{F}{m}$. Since both particles experience the same force $F$, but particle A has mass $m$ and particle B has mass $3m$, their accelerations differ. Particle A's acceleration is $\frac{F}{m}$, while particle B's is $\frac{F}{3m}$ — making A's acceleration three times larger. This confirms answer B. Choice A incorrectly claims the force on B is three times larger — a common mistake of confusing the charge ratio (1:3) with the force, ignoring Newton's Third Law. Choice C compounds this error by claiming A's force is larger and calculating a nine-times acceleration difference. Choice D correctly identifies equal forces but reverses the acceleration comparison — B has the larger mass, so it has the smaller acceleration, not larger. A useful mental anchor: same force, different mass → different acceleration. Equal and opposite forces are about the interaction, not the objects' properties.
15. A point charge of $+2.0 × 10^{-6} C$ and an unknown point charge are separated by $0.30 m$. They attract each other with a force of magnitude $0.72 N$. Using $k = 9.0 × 10^9 N·m^2/C^2$, what is the unknown charge?
    Answer: $-3.6 × 10^{-6} C$ When you see a question involving two charges, a distance, and a force, you're working with Coulomb's Law: $F = k\frac{|q_1||q_2|}{r^2}$. The key insight here is that this law gives you the magnitude of the force, but the sign of the unknown charge must be determined from the physical description. Start by solving for the unknown charge magnitude. Rearranging Coulomb's Law: $|q_2| = \frac{Fr^2}{k|q_1|}$. Plugging in the values: $|q_2| = \frac{(0.72)(0.30)^2}{(9.0 \times 10^9)(2.0 \times 10^{-6})} = \frac{(0.72)(0.09)}{1.8 \times 10^4} = \frac{0.0648}{18000} = 3.6 \times 10^{-6} \text{ C}$. So the magnitude is $3.6 \times 10^{-6}$ C. Now apply the sign: the problem says the charges attract each other. Opposite charges attract, and since the known charge is positive, the unknown must be negative. The correct answer is B, $-3.6 \times 10^{-6}$ C. A gets the magnitude right but assigns a positive sign — this would mean two positive charges, which repel, not attract. Don't ignore the word "attract." C results from an arithmetic error, likely forgetting to square the distance $r$, giving a smaller magnitude than correct. D comes from not squaring $r$ and mishandling the powers of ten in the calculation. A quick strategy: always handle magnitude and sign separately in Coulomb's Law problems. Compute the magnitude algebraically first, then use "opposites attract, likes repel" to assign the correct sign.

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