Physics 2 Quiz: Superposition Electric Force

What this quiz covers

This quiz focuses on Superposition Electric Force, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 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. (correct answer)
2. The claim is incorrect: the net force on the original $-Q$ was zero by symmetry, but adding $+Q$ at the center creates a dipole pair whose induced field breaks the symmetry and produces a nonzero net force on the combined charge.
3. The claim is correct only if $Q$ is sufficiently small so that the approximation of point charges remains valid; for finite-sized charges the force would no longer be zero due to overlap effects that violate the superposition principle.
4. The claim is incorrect: while the net force on the original $-Q$ was zero, the net force on the $+Q$ placed at the center is nonzero and directed outward, so the total force on the combined central system is nonzero.

Question 2

1. At $x = \frac{L}{1 - \sqrt{3}}$, which is located to the left of the $+3q$ charge (at $x < 0$), where the repulsion from $+3q$ and the attraction toward $-q$ are both directed in the same sense and therefore cannot produce a zero net force on a positive test charge.
2. At $x = \frac{L}{1 + \sqrt{3}}$, which is located between the two charges ($0 < x < L$), where the opposing forces from $+3q$ and $-q$ can balance if the distances are appropriately scaled.
3. 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 $\sqrt{3}:1$. (correct answer)
4. There is no position on the x-axis where the net force on $+q$ is zero, because the charge magnitudes differ by a factor of 3 and the attractive and repulsive forces always point in the same direction along the axis.

Question 3

1. The net force approaches $\frac{kQq}{d^2}\cdot\frac{\pi^2}{6}$ in the $+x$ direction, since the force is $kQq\sum_{n=1}^{N}\frac{1}{n^2 d^2}$ and the Basel sum $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ converges.
2. The net force diverges to infinity as $N \to \infty$ with fixed $Q$, because each additional charge added closer to the origin contributes an increasingly large Coulomb force that causes the partial sums to grow without bound.
3. The net force approaches $\frac{kQq}{(Nd)^2}$ in the $+x$ direction for large $N$, because at large $N$ the distributed charge behaves like a single point charge $Q$ located at the centroid $x = Nd/2$, which is far from the origin.
4. 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. (correct answer)

Question 4

1. The net force is zero, because the rectangle has a vertical line of symmetry at $x = 0$ that cancels all horizontal components, and a horizontal line of symmetry at $y = a/2$ that causes the vertical components from the top and bottom charges to cancel as well.
2. The net force points in the $-y$ direction (downward), because the two positive corner charges below the center repel it upward but the two negative corner charges above attract it upward — the bottom pair is closer to the center than the top pair, so the repulsive downward component from the asymmetry dominates.
3. The net force points in the $+y$ direction (upward), because the two negative charges above the center exert a stronger net attractive force than the two positive charges below exert a repulsive force, since the attractive Coulomb interaction between opposite charges scales more favorably with this rectangular geometry.
4. 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. (correct answer)

Question 5

1. 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. (correct answer)
2. The net force is directed toward the nearest corner charge and has magnitude $\frac{5k(5q)(q)}{r^2}$, where $r$ is the circumradius, because the central charge experiences the greatest attraction to whichever vertex charge is labeled first.
3. The net force is nonzero and directed outward away from the center of the pentagon, because the negative central charge is attracted to all five positive charges simultaneously and the resultant of five inward-directed attractive forces does not cancel for a regular pentagon.
4. The net force is zero only if $5q^2 = (5q)^2$, i.e., only when the product of the corner charge and the central charge equals the square of each individual charge, a condition that is not satisfied here, so the net force is nonzero.

Question 6

1. $\frac{4kq^2}{d^2}\sqrt{3}$, because both forces have the same magnitude $\frac{4kq^2}{d^2}$, and since one force is repulsive (from $+2q$ at bottom-left) and one is attractive (toward $-2q$ at bottom-right), the horizontal components cancel while the vertical components add constructively, giving a net vertical force scaled by $\sqrt{3}$.
2. $\frac{4kq^2}{d^2}\sqrt{2}$, because the repulsive force from the bottom-left charge and the attractive force from the bottom-right charge are perpendicular to each other as seen from the top vertex, and combining two equal-magnitude perpendicular vectors gives a resultant scaled by $\sqrt{2}$.
3. $\frac{4kq^2}{d^2}$, 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. (correct answer)
4. $\frac{8kq^2}{d^2}$, because the repulsion from $+2q$ and the attraction toward $-2q$ both contribute full-magnitude forces that add directly since both forces are directed horizontally toward the right from the perspective of the top vertex, doubling the single-pair force magnitude.

Question 7

1. The net force is approximately zero, because in the limit $h \to 0$ the third charge approaches the midpoint between the two fixed charges, and the repulsive force from $+Q$ at the origin and the attractive force from $-Q$ at $(L, 0)$ are equal in magnitude and point in exactly opposite directions, so they cancel.
2. 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. (correct answer)
3. The net force is directed in the $-y$ direction with magnitude approximately $\frac{8kQ^2 h}{L^3}$, because the horizontal force components cancel by symmetry and the small vertical displacement $h$ produces unequal distances to the two fixed charges, resulting in a net downward force proportional to $h$.
4. The net force is directed in the $+y$ direction with magnitude approximately $\frac{8kQ^2 h}{L^3}$, because both the repulsive force from $+Q$ at the origin and the attractive force from $-Q$ at $(L,0)$ have upward vertical components when the third charge is above the x-axis, producing a net upward force that vanishes as $h \to 0$.

Question 8

1. $\frac{3}{8}F_0$, 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 $\frac{k(Q/2)(3Q/4)}{R^2} = \frac{3}{8}F_0$. (correct answer)
2. $\frac{1}{4}F_0$, because each touching halves the charge on the sphere being touched, so sphere 1 ends up with $Q/2$ and sphere 2 ends up with $Q/2$, giving a force of $\frac{k(Q/2)(Q/2)}{R^2} = \frac{1}{4}F_0$.
3. $\frac{1}{2}F_0$, because the total charge $2Q$ is conserved across all three spheres and ultimately distributes so that spheres 1 and 2 together retain charge $Q$, each holding $Q/2$, but the force scales as $Q^2/2$ relative to the original.
4. $\frac{3}{16}F_0$, because the third sphere acts as a charge drain; after both contacts the charges on spheres 1 and 2 are reduced by successive halving and quartering, yielding a product of charges equal to $\frac{3}{16}Q^2$.

Question 9

1. The net force points in the $+x$ direction with magnitude $\frac{kq^2}{d^2}\left(4 - 1 - \frac{4}{9}\right)$, because the repulsion from $Q_3$ at distance $d$ is taken to act in the $+x$ direction and dominates over the attraction from $Q_2$ and the repulsion from $Q_1$, each at distances $d$ and $3d$ respectively.
2. The net force is zero, because the symmetric placement of $Q_1$ and $Q_3$ about the origin creates equal and opposite contributions to the net force on $Q_T$, and the remaining force from $Q_2$ is exactly balanced by the asymmetry in the distances from $Q_T$ to the two charges $Q_1$ and $Q_3$.
3. 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. (correct answer)
4. The net force points in the $-x$ direction, because the attraction from $Q_2$ at distance $d$ combines with the repulsion from $Q_3$ at distance $d$ to produce a large net force toward $-x$, with the contribution from the distant $Q_1$ at distance $3d$ being negligible and therefore ignored entirely in the calculation.