Question bank — Superposition principle for forces

If you double the charge $q_i$ of one source, the force it exerts on $q_0$ doubles and no other pair force changes. ::: True — Coulomb's law is linear in each source charge, and superposition says the other terms are untouched, so only that one term scales.

Placing a large charge between two interacting charges reduces the force between the original two. ::: False — in vacuum each pair force is independent; the middle charge only adds its own force term to each, it never screens or weakens an existing pairwise force.

Superposition is a logical consequence of Coulomb's inverse-square law. ::: False — inverse-square fixes each pair force, but the linearity (that forces simply add with no cross-terms) is a separate experimental fact that Coulomb's law happens to respect.

If two source forces on $q_0$ are equal in magnitude and opposite in direction, the net force is zero even though each source still acts. ::: True — the vectors cancel, but each charge is still exerting its full independent force; "zero net" is not "no forces."

Superposition holds for the electric field the same way it holds for force. ::: True — since $\vec{F}=q_0\vec{E}$ and $q_0$ is a common factor, $\vec{E}=\sum {\vec{E}}_i$ obeys the identical vector-sum rule (see Electric Field).

For a continuous charge (a charged rod), superposition no longer applies because there are no separate charges to add. ::: False — you split the rod into infinitesimal pieces $dq$, each a point source, and the sum becomes an integral $\vec{F}=\int d\vec{F}$ (see Continuous Charge Distributions).

Electric potential energy from several charges also requires vector addition. ::: False — potential (and potential energy) is a scalar, so you just add numbers with their signs — no direction, no components (see Electric Potential).

"Two forces of $3\text{N}$ and $4\text{N}$ act on a charge, so the net force is $7\text{N}$." ::: Only true if they are collinear and same direction. If perpendicular, the net is $\sqrt{3^2+4^2}=5\text{N}$; the answer depends on the angle, which was never stated.

"The unit vector ${\hat{r}}_{0i}$ for the force on $q_0$ points from $q_0$ toward the source $q_i$." ::: Backwards. For the force on $q_0$, ${\hat{r}}_{0i}$ points from the source $q_i$ toward $q_0$, so that like charges (positive product) automatically push $q_0$ away.

"$q_0$ is negative, so I put a minus sign in front of the final net force by hand." ::: No hand-added sign — the sign of $q_0$ lives inside each Coulomb term via the product $q_0{q}_i$. Fixing it twice double-counts and flips the direction wrongly.

"The net force came out as $(F_x,F_y)=(-2,-3)$, so the angle is $\arctan (3/2)\approx 56^{\circ }$ above the $+x$ axis." ::: Both components are negative, so the vector lives in the third quadrant; the true direction is $180^{\circ }+56^{\circ }\approx 236^{\circ }$. Bare $\arctan$ loses the quadrant — check the component signs (see Vector Addition and Resolution).

"I found each pair force's magnitude, added them to get $|\vec{F}|$, then found the angle from those magnitudes." ::: You cannot recover a direction from magnitudes alone. You must resolve each force into $x$ and $y$ components, sum components, then build magnitude and angle.

Why does the vector form of Coulomb's law carry $r^3$ in the denominator while the scalar form has $r^2$? ::: Writing the direction as $\hat{r}=({\vec{r}}_0-{\vec{r}}_i)/|{\vec{r}}_0-{\vec{r}}_i|$ adds one extra power of $r$ in the denominator; the numerator $({\vec{r}}_0-{\vec{r}}_i)$ then carries the direction.

Why could superposition, in principle, have been false? ::: If forces "saturated" (a charge could feel only so much total force) or if charges screened each other nonlinearly, the simple additive rule would break — nature just happens to be linear to high precision.

Why does knowing $q_0$'s sign let us drop the hand-drawn arrows for attraction vs repulsion? ::: The product $q_0{q}_i$ is positive for like signs (pushing $q_0$ outward along ${\hat{r}}_{0i}$) and negative for unlike signs (pulling inward), so the algebra sets the direction automatically.

Why is potential energy of a charge configuration easier to superpose than force? ::: Energy is a scalar — you add signed numbers with no direction — whereas force needs full vector bookkeeping across components.

Two identical positive charges sit symmetrically on either side of $q_0$ on a line. What is the net force? ::: Zero — the two forces are equal in magnitude and exactly opposite, so they cancel; $q_0$ is at an unstable equilibrium along that line.

What happens to a pair force as the separation $r\to 0$? ::: It diverges to infinity ($1/r^2\to \infty$); the point-charge model breaks down at contact, so this is an idealization limit, not a real infinite force.

As one source charge is moved infinitely far away ($r\to \infty$), what is its contribution to the net force? ::: It tends to zero like $1/r^2$, so distant charges drop out of the sum — but they never contribute a negative correction, they simply fade.

Three equal positive charges sit at the vertices of an equilateral triangle. Is the net force on any one of them zero? ::: No — the two forces are equal in magnitude but $60^{\circ }$ apart, so they add to a nonzero vector pointing radially outward from the triangle's center.

If all charges (source and target) are negative, does superposition change form? ::: No — the vector-sum formula is identical; the signs simply enter through each product $q_0{q}_i$, which is positive for these like charges, giving mutual repulsion.

A charge $q_0$ sits at the exact center of a uniformly charged ring. What net force does the ring exert? ::: Zero — by symmetry every $dq$ element has a diametrically opposite partner whose force cancels it, so the integrated superposition vanishes.