Exercises — Superposition principle for forces

What's wrong: $14\text{mN}$ is the scalar sum. But east and north are perpendicular directions — you cannot add their sizes directly, any more than you can add "3 steps forward" and "4 steps left" to get "7 steps."

What to do — look at the right triangle in the figure. The two arrows form the legs; the true total is the diagonal (the hypotenuse): $|\vec{F}|=\sqrt{6^2+8^2}\text{mN}=\sqrt{36+64}=\sqrt{100}=10\text{mN}$

No. Superposition says each pair-force is computed as if the others did not exist (see Coulomb's Law). The force ${\vec{F}}_{12}$ depends only on $q_1$, $q_2$, and their separation — there is no $q_3$ term in it.

What does change is the net force on $q_1$: it now also feels ${\vec{F}}_{13}$, a separate added arrow. The old ${\vec{F}}_{12}$ is untouched; a new one is stacked on top.

From $q_1$ (distance $1\text{m}$, like signs → repel → pushes $q_0$ away from $q_1$, i.e. $+x$): $F_{01}=\frac{k(2\times 10^{-6})(1\times 10^{-6})}{1^2}=9\times 10^9\cdot 2\times 10^{-12}=1.8\times 10^{-2}\text{N}(+x)$ From $q_2$ (distance $2\text{m}$, opposite signs → attract → pulls $q_0$ toward $q_2$ at $x=3$, i.e. $+x$): $F_{02}=\frac{k(3\times 10^{-6})(1\times 10^{-6})}{2^2}=\frac{9\times 10^9\cdot 3\times 10^{-12}}{4}=6.75\times 10^{-3}\text{N}(+x)$ Both point $+x$, so this is the one case where we can add sizes directly (they are collinear): $F_{\text{net}}=1.8\times 10^{-2}+6.75\times 10^{-3}=2.475\times 10^{-2}\text{N}=24.75\text{mN}\text{along }+x$

Why between them? Both source charges are positive, so a positive test charge is pushed right by $q_1$ and left by $q_2$. Only between them can these two pushes oppose and cancel. (Outside, both would point the same way — no cancellation.)

Let the test charge sit at distance $x$ from $q_1$, so it is $(3-x)$ from $q_2$. Cancellation means equal magnitudes (the common factors $k$ and the test charge cancel): $\frac{q_1}{x^2}=\frac{q_2}{(3-x{)}^2}\Rightarrow \frac{4}{x^2}=\frac{1}{(3-x{)}^2}$ Take the square root of both sides (both distances positive, so keep the $+$ root): $\frac{2}{x}=\frac{1}{3-x}\Rightarrow 2(3-x)=x\Rightarrow 6=3x\Rightarrow x=2\text{m}$ Answer: $2\text{m}$ from $q_1$ (i.e. $1\text{m}$ from $q_2$). The point sits closer to the weaker charge — sensible, since it must be near the small charge to make its push as strong as the distant big one.

Look at the figure — the two source charges are at two corners of a square, the target at a third.

From $q_1$ at origin: separation vector $(2,0)-(0,0)=(2,0)$, distance $2\text{m}$, unit vector $(1,0)$. Repulsion → along $+x$: $F_{01}=\frac{k(2)(1)\times 10^{-12}}{2^2}=\frac{9\times 10^9\cdot 2\times 10^{-12}}{4}=4.5\times 10^{-3}\text{N}\Rightarrow (4.5\times 10^{-3},0)$ From $q_2$ at $(0,2)$: separation vector $(2,0)-(0,2)=(2,-2)$, distance $\sqrt{2^2+2^2}=2\sqrt{2}\approx 2.828\text{m}$. Unit vector $=(2,-2)/(2\sqrt{2})=(0.707,-0.707)$ — this points from source $q_2$ toward target $q_0$, so repulsion pushes $q_0$ along it (down-and-right). $F_{02}=\frac{k(2)(1)\times 10^{-12}}{(2\sqrt{2}{)}^2}=\frac{18\times 10^{-3}}{8}=2.25\times 10^{-3}\text{N}$ ${\vec{F}}_{02}=2.25\times 10^{-3}(0.707,-0.707)=(1.591\times 10^{-3},-1.591\times 10^{-3})$ Add components (this is why we resolve — you cannot add sizes of arrows at $45^{\circ }$ to each other): $F_x=4.5\times 10^{-3}+1.591\times 10^{-3}=6.091\times 10^{-3}$ $F_y=0-1.591\times 10^{-3}=-1.591\times 10^{-3}$ Recombine: $|\vec{F}|=\sqrt{6.091^2+1.591^2}\times 10^{-3}=6.29\times 10^{-3}\text{N}=6.29\text{mN}$ $\varphi =\arctan \frac{-1.591}{6.091}=-14.65^{\circ }$ Answer: $\approx 6.29\text{mN}$, tilted $14.7^{\circ }$ below the $+x$ axis — the downward tilt is $q_2$'s pull-component. (See Vector Addition and Resolution for the machinery.)

Look at the figure — pick the top charge. The two lower charges each repel it; by symmetry their horizontal parts cancel and their vertical parts add.

Each pair force has magnitude $F=\frac{k{q}^2}{a^2}$ (all sides equal $a$).

Directions: each of the two lower charges pushes the top charge outward along the side. Each side makes $30^{\circ }$ with the vertical (since interior angles are $60^{\circ }$; the two arrows are symmetric about the vertical, splitting the $60^{\circ }$ apex angle into $30^{\circ }$ each side).

Horizontal parts: equal and opposite → cancel. ✅ (This is why we chose components — symmetry kills one axis.)

Vertical parts add: each contributes $F\cos 30^{\circ }$, and there are two: $F_{\text{net}}=2F\cos 30^{\circ }=2\cdot \frac{k{q}^2}{a^2}\cdot \frac{\sqrt{3}}{2}=\sqrt{3}\frac{k{q}^2}{a^2}$ Answer: $F_{\text{net}}=\sqrt{3}\frac{k{q}^2}{a^2}$, pointing straight away from the triangle's centre.

Set-up. The distance from centre to each corner is half the diagonal: $r=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}},r^2=\frac{a^2}{2}$ Each corner produces a force on $+Q$ of magnitude $F=\frac{kqQ}{r^2}=\frac{kqQ}{a^2/2}=\frac{2kqQ}{a^2}$ The trick — imagine all four were $+q$. Then the four repulsive forces point outward toward the four corners in perfectly balanced opposite pairs → net zero.

Our case is "all $+q$" plus a correction: one corner is $-q$ instead of $+q$. Changing $+q\to -q$ at one corner is the same as superposing an extra $-2q$ there (because $+q+(-2q)=-q$). By superposition we may treat these separately:

• The "all $+q$" arrangement → zero.
• The extra $-2q$ at one corner → attracts $+Q$ toward that corner with magnitude $F_{\text{net}}=\frac{k(2q)Q}{r^2}=\frac{k(2q)Q}{a^2/2}=\frac{4kqQ}{a^2}$ Answer: $F_{\text{net}}=\frac{4kqQ}{a^2}$, directed from the centre toward the negative corner. Superposition let us split a messy 4-body sum into "zero + one easy term."

Condition for zero tilt: the total $y$-component must vanish, $F_y=0$.

$q_1$ sits on the x-axis with the target → contributes no $y$-component. So all the $y$-force comes from $q_2$, and it can only be zero if $q_2=0$... unless we allow a fourth charge — but with just these two, $q_2$'s force always has a downward $y$-part $(-)$ as long as $q_2\ne 0$.

${\vec{F}}_{02}=\frac{k{q}_0{q}_2}{(L^2+d^2)}\cdot \frac{(L,-d)}{\sqrt{L^2+d^2}},F_{02,y}=-\frac{k{q}_0{q}_{2}d}{(L^2+d^2{)}^{3/2}}$ This is zero only if $q_2=0$ or $d=0$. So: Answer: there is no nonzero ratio that removes the tilt — with $q_1$ on-axis it never contributes vertically, so any off-axis $q_2$ leaves a residual tilt. The only escapes are $q_2=0$ (no second charge) or $d=0$ (both sources on the axis, making it collinear). This is a degenerate-case check: the naïve wish "tune the ratio" fails, and understanding why is the point.

Why an integral? The rod is not one charge but infinitely many infinitesimal charges. Superposition says the total force is the sum of every tiny piece's force — and a sum over a continuum is an integral (this is exactly the move in Continuous Charge Distributions).

Step 1 — chop the rod. Uniform charge means linear density $\lambda =Q/L$ (charge per metre). A slice of width $dx$ at distance $x$ from $q_0$ holds $dq=\lambda dx=\frac{Q}{L}dx$ Set up coordinates: $q_0$ at the origin, the rod running from $x=a$ (near end) to $x=a+L$ (far end).

Step 2 — force from one slice. Every slice lies on the same line as $q_0$, so all forces are collinear (no components needed — a mercy of the geometry): $dF=\frac{k{q}_{0}dq}{x^2}=\frac{k{q}_{0}Q}{L}\cdot \frac{dx}{x^2}$ Step 3 — superpose (integrate) from near end to far end: $F=\frac{k{q}_{0}Q}{L}{\int }_a^{a+L}\frac{dx}{x^2}=\frac{k{q}_{0}Q}{L}{\left[-\frac{1}{x}\right]}_a^{a+L}=\frac{k{q}_{0}Q}{L}\left(\frac{1}{a}-\frac{1}{a+L}\right)$ Combine the fractions: $\frac{1}{a}-\frac{1}{a+L}=\frac{(a+L)-a}{a(a+L)}=\frac{L}{a(a+L)}$ $\boxed{F=\frac{k{q}_{0}Q}{L}\cdot \frac{L}{a(a+L)}=\frac{k{q}_{0}Q}{a(a+L)}}$ Sanity check (limiting behaviour): if the rod is very short, $L\to 0$, then $a+L\to a$ and $F\to \frac{k{q}_{0}Q}{a^2}$ — exactly a point charge $Q$ at distance $a$. ✅ The formula collapses to the simplest case, as any correct derivation must.

$F=\frac{k{q}_{0}Q}{a(a+L)}=\frac{(9\times 10^9)(2\times 10^{-6})(5\times 10^{-6})}{0.5(0.5+1)}$ Numerator: $9\times 10^9\cdot 2\times 10^{-6}\cdot 5\times 10^{-6}=9\times 10^9\cdot 10^{-11}=9\times 10^{-2}$. Denominator: $0.5\times 1.5=0.75$. $F=\frac{9\times 10^{-2}}{0.75}=0.12\text{N}$ Answer: $F=0.12\text{N}$, directed away from the rod (like charges repel).

• L1 ::: recognising vector vs scalar, and non-interference of pairs.
• L2 ::: plugging Coulomb's law for collinear cases; the null-point equal-magnitude trick.
• L3 ::: resolving into components and reading the quadrant from signs.
• L4 ::: splitting a hard configuration via superposition ("zero + correction").
• L5 ::: turning the discrete sum into an integral for a continuous distribution.