Superposition: Electric Force - Physics 2

1. In the limit $h \to 0$, what is the direction and approximate magnitude of the net force on the third charge $+Q$ immediately after release?
   Answer: The net force is directed in the $+x$ direction with magnitude approximately $\frac{8kQ^2}{L^2}$, because the repulsion from $+Q$ at the origin pushes the third charge to the right, and the attraction toward $-Q$ at $(L, 0)$ also pulls the third charge to the right; both $x$-components act in the same direction and add, while the $y$-components cancel by symmetry. When a charge sits between two other charges, your first instinct might be to look for cancellation — but cancellation only occurs when forces point in opposite directions. Here, the geometry destroys that intuition entirely. Place the third charge $+Q$ at $(L/2, h)$. The fixed $+Q$ at the origin repels it, pushing it away from the origin — meaning the force points toward the upper-right, with a $+x$ component. The fixed $-Q$ at $(L, 0)$ attracts it, pulling it toward $(L,0)$ — meaning that force points toward the lower-right, also with a $+x$ component. Both horizontal components point in the same direction ($+x$) and add together. Meanwhile, the $y$-components are equal and opposite by the left-right symmetry of the configuration, so they cancel. As $h \to 0$, both charges are essentially at distance $L/2$ from the third charge, giving each force magnitude $\frac{kQ^2}{(L/2)^2} = \frac{4kQ^2}{L^2}$. The two $x$-components sum to $\frac{8kQ^2}{L^2}$, confirming B is correct. Choice A is the classic trap: students assume "midpoint = cancellation," forgetting that cancellation requires forces pointing in opposite directions. Here they don't — both horizontal components reinforce each other. Choice C incorrectly claims the $x$-components cancel by symmetry; they absolutely do not, since both forces have rightward horizontal components. Choice D correctly identifies that $y$-components don't cancel (they don't — wait, they do cancel), but then invents a net upward force; in reality the $y$-components are equal and opposite and vanish. Strategy tip: Before invoking symmetry cancellation, always draw a vector diagram and check whether the relevant components actually point in opposite directions — same-sign components add, they don't cancel.
2. At what position on the x-axis could the third charge $+q$ be placed so that it experiences zero net electric force? Assume $q > 0$ and $L > 0$.
   Answer: At $x = \frac{\sqrt{3}\,L}{\sqrt{3}-1}$, which is located to the right of $-q$ (at $x > L$), where the repulsion from $+3q$ and the attraction toward $-q$ point in opposite directions and balance with the required distance ratio of $\sqrt{3}:1$. When finding where a test charge experiences zero net force, your first task is to identify which region of the axis allows the forces to oppose each other. For a positive test charge $+q$ between a positive charge $+3q$ (at $x=0$) and a negative charge $-q$ (at $x=L$), both forces point in the same direction (the repulsion from $+3q$ pushes right, and the attraction toward $-q$ also pulls right), so equilibrium is impossible there. To the left of $+3q$, repulsion pushes left while attraction pulls right — again the same direction. Only to the right of $-q$ do the forces oppose: repulsion from $+3q$ pushes right, attraction toward $-q$ pulls left. Setting the magnitudes equal with the test charge at position $x > L$: $\frac{k(3q)(q)}{x^2} = \frac{k(q)(q)}{(x-L)^2}$ $3(x-L)^2 = x^2 \implies \sqrt{3}(x-L) = x \implies x = \frac{\sqrt{3}\,L}{\sqrt{3}-1}$ This confirms C is correct — the equilibrium point lies to the right of $-q$, with the required distance ratio of $\sqrt{3}:1$ from the two charges. A is wrong because it claims the left region produces equilibrium, but as shown above, both forces point in the same direction there. B is wrong for the same reason applied to the region between the charges. D is simply false — unequal charge magnitudes don't prevent equilibrium; they just shift where it occurs. Strategy tip: Always sketch the force directions in each region before doing algebra. Equilibrium requires forces in opposite directions, so eliminate regions geometrically first, then solve only in the valid region.
3. A physicist claims that adding a sixth charge $+Q$ directly on top of (at the same location as) the central charge $-Q$ will cause the net force on the central charge to remain zero. Which of the following best evaluates this claim?
   Answer: The claim is correct: placing $+Q$ at the center still yields a net force of zero on the combined central charge, because the four corner charges are symmetric and their contributions cancel by superposition regardless of the sign or magnitude of the central charge. When a question involves forces on a charge surrounded by a symmetric arrangement, your first instinct should be to apply the superposition principle: the net force on any charge equals the vector sum of all individual forces acting on it, completely independent of whatever else occupies that location. Here's the key insight: the four corner charges $+Q$ are arranged with perfect square symmetry around the center. By symmetry, the forces they exert on any charge sitting at the center must cancel — the force from each corner is exactly opposed by the force from the diagonally opposite corner. This cancellation holds regardless of the sign or magnitude of the central charge. Whether the central charge is $-Q$, $+Q$, or $-Q + Q = 0$, the four corner forces sum to zero. The central charge's value scales each force uniformly but never breaks the symmetry. So A is correct: the net force on the combined central charge remains zero. Choice B introduces a fictitious "dipole-induced field" that breaks symmetry — this is physically wrong. Two charges occupying the same point don't form a dipole (which requires separation), and no new asymmetry is introduced. Choice C incorrectly suggests superposition fails for finite-sized or overlapping charges. Superposition is a fundamental principle of electrostatics that doesn't break down this way — the question is about point charges, and even conceptually the overlap doesn't violate superposition. Choice D makes the subtle error of treating the $+Q$ as a separate object experiencing a nonzero force. The $+Q$ at the center does experience forces from the corners, but they cancel by the same symmetry argument — the net force on it is also zero. Your takeaway: symmetry + superposition = powerful tools. When charges are symmetrically arranged, the net force on anything at the center is zero, regardless of what that central charge is.
4. After both touchings, what is the net electric force between spheres 1 and 2, expressed in terms of the original force $F_0 = \frac{kQ^2}{R^2}$?
   Answer: $\frac{3}{8}F_0$, because after the first touch sphere 1 retains $Q/2$ and the third sphere carries $Q/2$; after the second touch sphere 2 and the third sphere each carry $\frac{Q/2 + Q}{2} = \frac{3Q}{4}$, giving a force of $\frac{k(Q/2)(3Q/4)}{R^2} = \frac{3}{8}F_0$. When identical conducting spheres touch, they share charge equally — this is the key principle being tested. Tracking charge step by step is essential here. First touch: The uncharged third sphere contacts sphere 1 (charge $+Q$). The total charge $Q$ splits equally, so sphere 1 retains $Q/2$ and the third sphere carries $Q/2$. Second touch: The third sphere (carrying $Q/2$) contacts sphere 2 (carrying $+Q$). The combined charge is $Q/2 + Q = 3Q/2$, which splits equally, leaving sphere 2 with $3Q/4$ and the third sphere with $3Q/4$. Final force: Spheres 1 and 2 carry $Q/2$ and $3Q/4$ respectively, so: $F = \frac{k(Q/2)(3Q/4)}{R^2} = \frac{3}{8}\cdot\frac{kQ^2}{R^2} = \frac{3}{8}F_0$ This confirms answer A is correct. Answer B incorrectly assumes both spheres get halved independently — it ignores that sphere 2 still has its original charge $Q$ when the third sphere (now carrying $Q/2$) touches it, so the split isn't simply $Q/2$. Answer C invokes charge conservation loosely but applies faulty reasoning about distribution; the total across all three spheres is $2Q$, but how it partitions matters, and sphere 1's charge isn't affected by the second touch at all. Answer D invents a "charge drain" narrative with no physical basis — the third sphere redistributes charge, it doesn't destroy it. Study tip: Always track each charge-sharing event sequentially and independently. Write down the charge on every sphere before and after each touch — careless bookkeeping is exactly what these distractors exploit.
5. Five equal positive charges $+q$ are placed at the vertices of a regular pentagon. A sixth charge $-5q$ is placed at the geometric center. Which of the following correctly describes the net electric force on the $-5q$ charge?
   Answer: The net force is zero, because the five corner charges are symmetrically arranged and their individual Coulomb force vectors on the central charge sum to zero by vector cancellation, regardless of the magnitude of the central charge. Whenever you see a charge placed at the center of a symmetric arrangement of other charges, your first instinct should be to analyze the symmetry of the force vectors, not the magnitudes involved. For a regular pentagon, the five corner charges are equally spaced at identical distances $r$ from the center. Each $+q$ charge exerts an attractive Coulomb force on the central $-5q$ charge directed inward (toward that corner). Because the pentagon has perfect 5-fold rotational symmetry, these five force vectors are evenly distributed in angle — separated by $72°$ each. When you add five vectors of equal magnitude that point symmetrically in all directions around a full rotation, they cancel exactly. This is the same reason a charge at the center of any regular polygon experiences zero net force from equal charges at the vertices. The magnitude of the central charge is completely irrelevant to this cancellation — only the symmetry matters. So A is correct. B is wrong on two counts: there is no physical reason to favor a "first labeled" vertex, and the force formula shown incorrectly multiplies the distance squared as if all five charges act in the same direction. C contains a subtle but critical error — you're right that each force is directed inward (attractive), but five symmetrically inward-pointing vectors do cancel for a regular pentagon. The claim that they don't is false. D invents a nonsensical algebraic condition ($5q^2 = (5q)^2$) that has no basis in physics — symmetry, not charge algebra, determines cancellation. Your strategy: when charges are arranged with regular geometric symmetry, draw the force vectors and check for cancellation before doing any calculation. Symmetry almost always gives you the answer faster.
6. A fifth charge $+q$ is placed at the center of the rectangle at $(0, a/2)$. What is the direction of the net electric force on this central charge?
   Answer: The net force points in the $+y$ direction (upward), because all four corner charges are equidistant from the center, so left-right symmetry cancels all $x$-components; the positive charges below the center repel the central charge upward and the negative charges above attract it upward, so all four $y$-components point in the $+y$ direction and add constructively. When analyzing electric forces on a charge surrounded by a symmetric arrangement, your first move should always be to identify symmetry axes that cancel components — then carefully determine the direction of the remaining net force from each charge. Here, the rectangle has a vertical line of symmetry at $x = 0$. Because the four corner charges are arranged symmetrically left-right, every horizontal ($x$) force component cancels perfectly. That leaves only vertical ($y$) components to consider. Now check the distances. The center is at $(0, a/2)$. The bottom corners $(\pm a, 0)$ are each a distance $\sqrt{a^2 + (a/2)^2} = \frac{a\sqrt{5}}{2}$ away. The top corners $(\pm a, a)$ are the same distance: $\sqrt{a^2 + (a/2)^2} = \frac{a\sqrt{5}}{2}$. All four charges are equidistant from the center. This is the crucial geometric insight. Since all four are equidistant, force magnitudes are equal. The two positive charges below repel the central $+q$ — pushing it upward $(+y)$. The two negative charges above attract the central $+q$ — pulling it upward $(+y)$. All four vertical components point in the same direction and add together, making D correct. Answer A fails because it wrongly assumes a horizontal symmetry at $y = a/2$ cancels vertical forces — it would if the top and bottom charges were identical, but they have opposite signs, so their vertical effects reinforce rather than cancel. Answer B incorrectly claims the bottom pair is closer, which geometry disproves. Answer C reaches the right direction but invents a false justification about Coulomb scaling rather than using the actual equidistance argument. Your strategy: always compute distances before assuming which charges dominate. Symmetry cancels components; distance and sign determine what remains.
7. What is the magnitude of the net electric force on the charge at the top vertex due to the other two charges?
   Answer: $\frac{4kq^2}{d^2}$, because the two forces on the top charge have equal magnitude $\frac{4kq^2}{d^2}$ and their vertical (y) components cancel while their horizontal (x) components are equal and in the same direction, giving a net horizontal force equal in magnitude to a single force component. When tackling electric force problems with multiple charges, your first move should always be to draw the geometry carefully and decompose each force vector into components before combining them. In an equilateral triangle, every interior angle is 60°. From the top vertex, the bottom-left charge sits 60° below the horizontal to the left, and the bottom-right charge sits 60° below the horizontal to the right. Both charges are distance $d$ away, and both have magnitude $2q$, so both forces on the top charge have the same magnitude: $F = \frac{k(2q)(2q)}{d^2} = \frac{4kq^2}{d^2}$. Now consider directions. The bottom-left charge is $+2q$, so it repels the top $+2q$ charge — the force points upper-left (away from bottom-left). The bottom-right charge is $-2q$, so it attracts the top charge — the force points lower-right (toward bottom-right). Both forces therefore point to the right in the horizontal direction, while one points upward and the other downward in the vertical direction. The vertical components are equal and opposite, so they cancel. The horizontal components are both $F\cos(60°) = \frac{4kq^2}{d^2} \cdot \frac{1}{2} = \frac{2kq^2}{d^2}$, and they add: $F_{net} = \frac{4kq^2}{d^2}$. This confirms C. Choice A incorrectly claims the horizontal components cancel — it has the geometry backwards. Choice B assumes the two forces are perpendicular to each other, which is only true if the angle between them is 90°; here it's 60°. Choice D claims both forces point fully horizontally, ignoring that 60° below horizontal has a significant vertical component. Strategy tip: Always resolve forces into x- and y-components before adding. Never assume forces point in the same direction just because they act on the same charge — direction depends on both the sign of the charges and the geometry.
8. For large $N$, which expression best approximates the net force on the charge $-q$ at the origin due to all $N$ positive charges, assuming the total charge $Nq = Q$ remains fixed as $N \to \infty$?
   Answer: The net force approaches zero as $N \to \infty$ with fixed $Q$, because each individual charge $q = Q/N$ becomes infinitesimally small and the force from each vanishes, while the sum of vanishing forces also vanishes. When you see a question involving a sum of forces where both the number of charges and the charge per particle are changing simultaneously, you need to track how each term scales before worrying about the sum. Here, each positive charge has magnitude $q = Q/N$, placed at position $x = nd$. The force on $-q$ from the charge at position $nd$ is: $F_n = \frac{k(Q/N)(Q/N)}{(nd)^2}$ Wait — let's be careful. The force on $-q$ from one positive charge $q = Q/N$ at distance $nd$ is: $F_n = \frac{k \cdot (Q/N) \cdot (Q/N)}{(nd)^2}$ No — the charge at the origin is $-q = -Q/N$, and it also scales with $N$. So every individual force term scales as $1/N^2$, and summing $N$ such terms gives a total scaling as $N \cdot (1/N^2) = 1/N \to 0$. This confirms D: as $N \to \infty$ with fixed $Q$, the net force vanishes. A is tempting but contains a hidden error: it treats $q$ in the force formula as the fixed charge $Q$, not the shrinking $Q/N$. The Basel sum is mathematically real, but the coefficients in front are $\sim 1/N^2$, collapsing the result to zero. B claims divergence, which would require adding charges closer to the origin — but the minimum distance here is always $d$, so no charge ever approaches zero distance. C incorrectly applies a point-charge approximation at the centroid; that method estimates the field far from a distribution, not the force within it. The key strategy: when both $N$ and $q$ are changing, write every quantity explicitly in terms of $N$ before summing — never assume one change dominates without checking the full scaling.
9. What is the direction and relative magnitude of the net electric force on $Q_T$?
   Answer: The net force points in the $-x$ direction, because the attraction toward $Q_2$ at distance $d$ and the repulsion from $Q_3$ at distance $d$ both push $Q_T$ in the $-x$ direction, and together they outweigh the repulsion from $Q_1$ at distance $3d$, giving a net magnitude of $\frac{kq^2}{d^2}\left(1 + 4 - \frac{4}{9}\right)$ in the negative direction. When multiple charges act on a single test charge, your job is to calculate the force from each charge individually — including its direction — and then sum them as vectors. The sign of each source charge and the position of the test charge both matter for determining direction. Here, $Q_T = +q$ sits at $x = +d$. Let's identify each force: - **From $Q_2 = -q$ at $x = 0$** (distance $d$): opposite charges attract, so the force pulls $Q_T$ in the $-x$ direction. Magnitude: $\frac{kq^2}{d^2}(1)$ - **From $Q_3 = +4q$ at $x = +2d$** (distance $d$): same-sign charges repel, pushing $Q_T$ away from $Q_3$, i.e., in the $-x$ direction. Magnitude: $\frac{kq^2}{d^2}(4)$ - **From $Q_1 = +4q$ at $x = -2d$** (distance $3d$): same-sign charges repel, pushing $Q_T$ in the $+x$ direction. Magnitude: $\frac{kq^2}{d^2}\left(\frac{4}{9}\right)$ Net force: $\frac{kq^2}{d^2}\left(1 + 4 - \frac{4}{9}\right)$ in the $-x$ direction. This confirms C. A is wrong because it incorrectly assigns the repulsion from $Q_3$ as pointing in the $+x$ direction — a critical sign error. $Q_3$ is to the right of $Q_T$, so repulsion pushes $Q_T$ left. B is wrong because $Q_1$ and $Q_3$ are not equidistant from $Q_T$ — they are at distances $3d$ and $d$ respectively, so their forces do not cancel. D contains correct directional reasoning but makes an unjustified physical approximation by ignoring $Q_1$ entirely. On an exam, always include all contributions unless explicitly told a charge is negligible. The strategy to remember: always establish direction before calculating magnitude for each force. A common trap is misidentifying which way repulsion acts when the repelling charge is closer than you expect.

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