Coulomb's Law - Physics 2

1. Two point charges $q_A$ and $q_B$ are separated by a distance $r$. The force on $q_A$ due to $q_B$ has magnitude $F$. The charge $q_B$ is now doubled AND the separation is increased by a factor of $\sqrt{2}$. Simultaneously, a third charge $q_C = q_B$ (the original value of $q_B$) is placed at the same location as the new position of $q_B$. What is the net force magnitude on $q_A$ due to the charges at the new location?
   Answer: $\frac{3F}{2}$, because the total charge at the new location is $2q_B + q_B = 3q_B$, the new distance is $r\sqrt{2}$ so the distance-squared factor becomes $2r^2$, and thus the new force is $\frac{k q_A (3q_B)}{2r^2} = \frac{3}{2} \cdot \frac{kq_Aq_B}{r^2} = \frac{3F}{2}$. Whenever you see a Coulomb's law problem with multiple simultaneous changes, your instinct should be to write out the full expression and substitute carefully — never try to track the changes mentally. Start with what you know: the original force is $F = \frac{kq_Aq_B}{r^2}$. Now apply all three changes at once. The charge $q_B$ is doubled to $2q_B$, and a second charge $q_C = q_B$ is placed at the same new location. Because both charges sit at identical positions, superposition tells you to simply add them: the total charge at the new location is $2q_B + q_B = 3q_B$. The separation increases to $r\sqrt{2}$, so the denominator becomes $(r\sqrt{2})^2 = 2r^2$. The new force is therefore: $F' = \frac{kq_A(3q_B)}{2r^2} = \frac{3}{2}\cdot\frac{kq_Aq_B}{r^2} = \frac{3F}{2}$ This confirms D is correct. Choice A is wrong because it claims the effects "cancel out" and restore $F$ — a classic trap of sloppy intuition. The numerator grows by a factor of 3 while the denominator only grows by 2, so they do not cancel. Choice B correctly identifies the $3q_B$ total charge and the factor of 2 in the denominator, arriving at $\frac{3F}{2}$ — but then contradicts itself by claiming this "approximately" equals $2F$, which is unjustified. Choice C introduces a spurious $\sqrt{2}$ factor; since both charges are at the same location, you add them before applying Coulomb's law — no separate vector decomposition is needed. Your go-to strategy: always write $F = \frac{kq_1q_2}{r^2}$, substitute the new values explicitly, and simplify. Never eyeball combined changes.
2. Two point charges, $q_1 = +3\,\mu\text{C}$ and $q_2 = -12\,\mu\text{C}$, are separated by a distance of $0.6\,\text{m}$ in vacuum. A third charge $q_3$ is placed along the line connecting them such that the net force on $q_3$ is zero. Ignoring the forces on $q_1$ and $q_2$ themselves, at what distance from $q_1$ must $q_3$ be placed?
   Answer: $0.20\,\text{m}$ from $q_1$, placing $q_3$ between the two charges, where the $1/r^2$ dependence compensates for the 4:1 charge magnitude ratio. When a third charge is placed along the line connecting two charges and experiences zero net force, the forces from each charge must be equal in magnitude and opposite in direction. This requires careful thinking about both geometry (where along the line) and algebra (setting force magnitudes equal). Start by identifying the region. Since $q_1 = +3\,\mu\text{C}$ and $q_2 = -12\,\mu\text{C}$ have opposite signs, they attract any third charge in the same direction when $q_3$ sits between them — meaning the forces on $q_3$ can point in opposite directions and potentially cancel. Let $d$ be the distance from $q_1$, so the distance from $q_2$ is $(0.6 - d)$. Setting the Coulomb force magnitudes equal: $\frac{k|q_1|}{d^2} = \frac{k|q_2|}{(0.6-d)^2}$ $\frac{3}{d^2} = \frac{12}{(0.6-d)^2} \implies (0.6-d)^2 = 4d^2 \implies 0.6 - d = 2d \implies d = 0.20\,\text{m}$ This confirms answer A is correct — $q_3$ sits 0.20 m from $q_1$, between the charges, where the $1/r^2$ dependence compensates for the 4:1 charge ratio. Answer B is wrong because the midpoint (0.30 m) doesn't satisfy the force-balance equation — equal distances would require equal charges. Answer C places $q_3$ beyond $q_2$ (exterior region), but for opposite-sign charges, both forces on $q_3$ in the exterior region beyond $q_2$ point in the same direction — they cannot cancel. Answer D applies similar flawed reasoning to the exterior region beyond $q_1$; again, both forces align rather than oppose. A quick strategy: opposite-sign charges → balance point is between them; same-sign charges → balance point is outside, closer to the weaker charge. Memorize this rule before the exam.
3. An experimenter measures the Coulomb force between two point charges in vacuum and finds it to be $F_0$. The experimenter then makes three simultaneous changes: (i) the sign of one charge is reversed, (ii) the magnitude of each charge is tripled, and (iii) the separation is tripled. What is the magnitude of the new force, and does it become attractive or repulsive?
   Answer: Magnitude $F_0$, attractive, because tripling both charges gives a factor of $9$ in the numerator, tripling the distance gives a factor of $9$ in the denominator, and the sign reversal changes the interaction from repulsive to attractive with no effect on magnitude. When multiple simultaneous changes are made to a Coulomb force problem, your strategy should be to track each change as a separate multiplicative factor, then combine them — and handle the magnitude and sign (attractive vs. repulsive) as two independent questions. Coulomb's law states $F = k\frac{|q_1||q_2|}{r^2}$. Starting from $F_0$, here's what each change does: tripling both charge magnitudes multiplies the numerator by $3 \times 3 = 9$; tripling the separation multiplies the denominator by $3^2 = 9$. The net factor is $\frac{9}{9} = 1$, so the new magnitude equals $F_0$. Separately, reversing the sign of one charge means the two charges now have opposite signs, making the force attractive. This is answer A — magnitude $F_0$, attractive. B is wrong because it correctly computes the magnitude as $F_0$ but then claims the sign reversal doesn't affect whether the force is attractive or repulsive. It absolutely does — opposite signs always produce attraction. C commits a fundamental algebra error, claiming each charge "independently triples the force" for a factor of 6. That's not how multiplication works in the numerator: $3q_1 \times 3q_2 = 9q_1q_2$, a factor of 9, not 6. D invents a fictional "factor of 9 from field orientation" due to the sign reversal. Sign reversal only determines the direction (attractive or repulsive) — it contributes no numerical factor to the magnitude whatsoever. Your study tip: always separate a Coulomb problem into two clean steps — compute the ratio of magnitudes using only absolute values, then determine attractive vs. repulsive purely from whether the signs are like or opposite.
4. Two identical conducting spheres, each carrying charge $+Q$, are held a distance $d$ apart and experience a repulsive force $F_0$. The spheres are briefly brought into contact with each other and then separated back to the same distance $d$. One sphere is then touched to a large grounded conductor, removing all its charge, and the two spheres are again held at distance $d$. What is the magnitude of the new force between them, expressed in terms of $F_0$?
   Answer: $0$, because one sphere carries no charge after grounding, and by Coulomb's law the force between a charged object and an uncharged object is identically zero for point charges, regardless of the charge on the other sphere. When you see a question involving charge redistribution and Coulomb's law, work through each step sequentially — track what charge each sphere actually carries before applying any force formula. Here's the chain of events: Initially, both spheres carry $+Q$, producing force $F_0 = k\frac{Q \cdot Q}{d^2}$. When the identical spheres touch, the total charge $2Q$ distributes equally, leaving each with $+Q/2$. When one sphere is then touched to a large grounded conductor, all its charge drains away — it becomes completely neutral, carrying exactly zero charge. Now you have one sphere with $+Q/2$ and one sphere with $0$ charge. Coulomb's law is $F = k\frac{q_1 q_2}{d^2}$. If either charge is zero, the product $q_1 q_2 = 0$, and therefore the force is exactly zero. This confirms C is correct. Choice A fabricates a chain of fractional reductions that has no physical basis — you cannot get a nonzero force when one charge is zero, no matter how you scale it. Choice B invokes "image-charge effects" on the grounded sphere, but that phenomenon applies only when the sphere remains connected to ground while near a charged object, creating an induced charge. Here, the sphere is simply touched to ground and then separated — it leaves neutral, with no ongoing induction. Choice D misreads "removing all charge" as removing only $Q$, but grounding removes whatever charge is present, leaving zero — not $+Q/2 - Q$. As a study rule: always apply Coulomb's law to the final charges, not intermediate states, and remember that a grounded-then-separated conductor departs with zero net charge.
5. Four point charges, each of magnitude $q$, are placed at the corners of a square with side length $a$. Two adjacent corners carry charge $+q$ and the other two adjacent corners carry charge $-q$, so that the charges alternate in sign around the square. What is the magnitude of the net Coulomb force on one of the $+q$ charges?
   Answer: $\frac{kq^2}{a^2}\left(\sqrt{2} - \frac{1}{2}\right)$, because the two adjacent $-q$ charges each exert an attractive force of magnitude $kq^2/a^2$ along the sides; their vector resultant points diagonally with magnitude $\sqrt{2}\,kq^2/a^2$, and the repulsive force from the $+q$ at the opposite corner has magnitude $kq^2/(2a^2)$ and opposes this resultant. When you see a Coulomb force problem with multiple charges, your first move should always be to identify each pairwise force as a vector — magnitude and direction — then add them as components. Skipping the vector step is where most mistakes happen here. Focus on one of the $+q$ charges, say the one at the top-left corner. Label the square so that the two adjacent corners (top-right and bottom-left) each hold $-q$, and the opposite corner (bottom-right) holds $+q$. The two $-q$ neighbors each exert an attractive force of magnitude $kq^2/a^2$ directed along the sides — one pointing right, one pointing down. These two equal perpendicular forces combine by the Pythagorean theorem into a resultant of $\sqrt{2}\,kq^2/a^2$, directed diagonally toward the opposite corner. The $+q$ charge at that opposite corner is separated by a distance of $a\sqrt{2}$, so it exerts a repulsive force of $kq^2/(a\sqrt{2})^2 = kq^2/(2a^2)$ directed away from that corner — exactly opposing the attractive resultant. Subtracting gives a net magnitude of $\frac{kq^2}{a^2}\!\left(\sqrt{2} - \frac{1}{2}\right)$, confirming C is correct. A is wrong because it incorrectly claims the two side forces add along the diagonal to give $2\sqrt{2}\,kq^2/a^2$, ignoring that vector addition of two perpendicular equal forces yields $\sqrt{2}$, not $2\sqrt{2}$. B gets the attractive resultant right but then says the repulsive diagonal force adds to it — it actually opposes it, since the repulsion pushes away from the opposite corner while the attractive resultant pulls toward it. D double-counts incorrectly and arrives at an implausibly simple cancellation. A reliable strategy: always draw the square, label each force arrow explicitly, and resolve into components before combining. The most common trap here is forgetting that "toward the opposite corner" and "away from the opposite corner" are antiparallel — meaning you subtract, not add.
6. The force between two point charges separated by distance $d$ in a medium of dielectric constant $\kappa = 4$ is measured to be $F_0$. The medium is then replaced with vacuum ($\kappa = 1$), and the separation is simultaneously halved to $d/2$. What is the new force between the charges?
   Answer: $16F_0$, because the dielectric constant of 4 suppressed the original force by a factor of 4, and halving the separation increases the force by a factor of 4 via the inverse-square law, so the two multiplicative factors of 4 combine to give a total factor of 16. Whenever you see a question combining a dielectric medium with a changing separation distance, treat each effect as a separate multiplicative factor and apply them together at the end. Start with Coulomb's Law in a dielectric medium: $F = \frac{kq_1 q_2}{\kappa d^2}$. The dielectric constant $\kappa$ appears in the denominator, meaning the medium suppresses the force. So the original force is $F_0 = \frac{kq_1 q_2}{4d^2}$. Now apply both changes simultaneously. Replacing the medium with vacuum sets $\kappa = 1$, removing the suppression — this multiplies the force by 4. Halving the distance to $d/2$ means $d^2$ becomes $(d/2)^2 = d^2/4$, so the force increases by a factor of 4 via the inverse-square law. Combined: $F_{\text{new}} = 4 \times 4 \times F_0 = 16F_0$. Answer B is correct. A is wrong because it claims the two effects "partially offset each other" — they don't. Both effects increase the force, so they multiply together, not cancel. C introduces a fictitious "polarization correction" that doesn't exist in standard Coulomb's Law; there's no subtraction step. D confuses the direction of both effects, suggesting the dielectric factor "offsets" the distance factor. In reality, removing a dielectric and decreasing separation both work in the same direction — both increase the force. A useful strategy: always write out $F = \frac{kq_1 q_2}{\kappa d^2}$ explicitly and track how each variable in the formula changes. Changes in opposite directions cancel; changes in the same direction compound.
7. A charge $+Q$ is fixed at the origin. A particle of mass $m$ and charge $-q$ (where $q > 0$) is released from rest at position $(d, 0)$. A second particle of mass $m$ and charge $-q$ is simultaneously released from rest at position $(0, d)$. Treating only the Coulomb interaction between each negative charge and the fixed $+Q$, and ignoring gravity as well as any interaction between the two $-q$ charges, which statement correctly compares the magnitudes of the initial accelerations of the two $-q$ particles?
   Answer: Both particles have identical initial accelerations of magnitude $\frac{kQq}{md^2}$, because both are at the same distance $d$ from $+Q$, carry the same charge magnitude $q$, and have the same mass $m$, so Coulomb's law gives the same force magnitude and Newton's second law gives the same acceleration magnitude. When comparing accelerations of charged particles, your first instinct should be to apply Coulomb's law and Newton's second law directly — and resist the temptation to overcomplicate the geometry. Coulomb's law gives the force magnitude between two charges separated by distance $r$ as $F = \frac{kQq}{r^2}$. Both $-q$ particles sit exactly distance $d$ from the origin (one at $(d,0)$, one at $(0,d)$), carry the same charge magnitude $q$, and interact with the same fixed charge $+Q$. Therefore, both experience identical force magnitudes: $F = \frac{kQq}{d^2}$. Since both have mass $m$, Newton's second law gives both the same acceleration magnitude: $a = \frac{kQq}{md^2}$. Answer D is correct. The distractors all fall into the same trap: treating the direction of the force as if it affects the magnitude of the acceleration. Answer A incorrectly claims the particle at $(0,d)$ has a smaller "effective component" — but the entire Coulomb force on each particle already points directly toward the origin (the attractive direction), with no component to resolve away. Answer B makes the opposite error, wrongly privileging the y-axis as somehow more "efficient." Answer C invents a $\sqrt{2}$ factor from a misapplied vector decomposition — neither force requires decomposition since both already point purely radially toward the origin. The key study tip: Coulomb's law depends only on distance and charge magnitudes, not on which axis a particle sits along. Direction determines where the particle moves next, not how hard it's initially pushed.

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