Exercises — Coulomb's law — force, comparison with gravity

The rule: like charges repel, unlike charges attract. Multiply the signs.

• (a) $(+)(+)=+$ → repel.
• (b) $(-)(+)=-$ → attract.
• (c) $(-)(-)=+$ → repel. Notice we never touched the formula — sign of the product alone decides direction.

Neither — the forces are equal in magnitude. Recall our subscript convention: ${\vec{F}}_{12}$ is the force on charge 1 due to charge 2, and ${\vec{F}}_{21}$ is the force on charge 2 due to charge 1. By Newton's third law, ${\vec{F}}_{12}=-{\vec{F}}_{21}$: the push on charge 1 by charge 2 is exactly as big as the push on charge 2 by charge 1. The formula $F=k|q_1{q}_2|/r^2$ uses the product $q_1{q}_2$, which is the same no matter which charge you call "the one feeling the force." A big charge and a small charge feel the same size push — the small one just accelerates more (that's mass, not force).

Step 1 — SI units. $q_1=5\times 10^{-6}\text{C}$, $q_2=2\times 10^{-6}\text{C}$, $r=0.20\text{m}$. Why: the formula is calibrated for coulombs and metres.

Step 2 — plug into the magnitude form. $F=(8.99\times 10^9)\frac{(5\times 10^{-6})(2\times 10^{-6})}{(0.20{)}^2}.$

Step 3 — arithmetic. Numerator of the fraction: $(5\times 10^{-6})(2\times 10^{-6})=10\times 10^{-12}=1.0\times 10^{-11}$. Denominator: $(0.20{)}^2=0.04$. $F=8.99\times 10^9\times \frac{1.0\times 10^{-11}}{0.04}=8.99\times 10^9\times 2.5\times 10^{-10}=2.25\text{N}.$

Answer: $F\approx 2.25\text{N}$, repulsive (both positive).

Step 1. Magnitudes: $|q_1|=|q_2|=1.6\times 10^{-19}\text{C}$.

Step 2. $F=(8.99\times 10^9)\frac{(1.6\times 10^{-19}{)}^2}{(5.3\times 10^{-11}{)}^2}.$

Step 3. Numerator: $(1.6\times 10^{-19}{)}^2=2.56\times 10^{-38}$. Denominator: $(5.3\times 10^{-11}{)}^2=2.809\times 10^{-21}$. $F=8.99\times 10^9\times \frac{2.56\times 10^{-38}}{2.809\times 10^{-21}}\approx 8.2\times 10^{-8}\text{N}.$

Answer: $F\approx 8.2\times 10^{-8}\text{N}$, attractive (opposite signs). Tiny in everyday terms, but it holds the atom together.

Step 1 — list the pushes on $q_B$. Two forces act: from $A$ (on the left) and from $C$ (on the right). Both are positive-positive → repulsive. In the figure these are the two red arrows.

• $A$ repels $B$ → pushes $B$ to the right ($+x$).
• $C$ repels $B$ → pushes $B$ to the left ($-x$).

Step 2 — sizes. Both source charges are $3\mu \text{C}$ at the same distance $0.10\text{m}$ from $B$: $F_{AB}=F_{CB}=(8.99\times 10^9)\frac{(3\times 10^{-6}{)}^2}{(0.10{)}^2}=(8.99\times 10^9)\frac{9\times 10^{-12}}{0.01}=8.09\text{N}.$

Step 3 — vector sum. They are equal in size but opposite in direction (the two equal-length red arrows), so they cancel: $F_{\text{net}}=+8.09-8.09=0\text{N}.$

Answer: the middle charge feels zero net force — it sits at a symmetry point. (It is unstable along the line, though: nudge it and one force wins.)

Step 1 — directions flip. Now $A$ and $B$ are unlike → attract; $B$ is pulled toward $A$, i.e. to the left. Likewise $C$ attracts $B$ to the right. (In the figure, mentally reverse both red arrows.) Step 2 — sizes unchanged. Magnitude uses $|q_1{q}_2|$, and $|-3|=3$, so each force is still $8.09\text{N}$. Step 3 — sum. Again equal and opposite: $F_{\text{net}}=-8.09+8.09=0\text{N}.$ Answer: still zero. Flipping $B$'s sign reversed both forces together, so their cancellation survives. This time the equilibrium is stable along the line (a nudge left increases the rightward pull back toward centre).

Step 1 — geometry. $Q$ is above the midpoint, so it is the same distance from each of the two lower charges. Horizontal offset to each $=0.15\text{m}$, vertical offset $=0.20\text{m}$. $r=\sqrt{0.15^2+0.20^2}=\sqrt{0.0225+0.04}=\sqrt{0.0625}=0.25\text{m}.$

Step 2 — size of each force. $F=(8.99\times 10^9)\frac{(4\times 10^{-6})(1\times 10^{-6})}{(0.25{)}^2}=(8.99\times 10^9)\frac{4\times 10^{-12}}{0.0625}=0.575\text{N}.$ Each lower charge repels $Q$ (all positive), pushing it up and outward along the line from that charge to $Q$.

Step 3 — split into components. Why: to add two arrows pointing in different directions, break each into horizontal ($x$) and vertical ($y$) pieces, add like-with-like. The direction cosines from the geometry: horizontal fraction $=0.15/0.25=0.6$, vertical fraction $=0.20/0.25=0.8$.

• Left charge pushes $Q$ up-and-right: $F_x=+0.575(0.6)=+0.345$, $F_y=+0.575(0.8)=+0.460$.
• Right charge pushes $Q$ up-and-left: $F_x=-0.345$, $F_y=+0.460$.

Step 4 — add. $F_{x,\text{net}}=+0.345-0.345=0,F_{y,\text{net}}=0.460+0.460=0.920\text{N}.$ By symmetry the horizontal parts cancel and the vertical parts reinforce.

Answer: $F_{\text{net}}\approx 0.92\text{N}$, pointing straight up ($+y$), away from the pair.

Step 1 — where can it balance? Call the test charge $q_3$ at position $x$. For the two repulsions (or two attractions — same reasoning) to cancel, they must point opposite ways and be equal in size.

• Between the charges ($0<x<0.50$): $q_1$ pushes right, $q_2$ pushes left — they oppose, so cancellation is possible here (this is the region shown in the figure).
• Outside (left of $q_1$ or right of $q_2$): both push the same way, cannot cancel.

So look between them.

Step 2 — set magnitudes equal. Let the distance from $q_1$ be $x$, so the distance from $q_2$ is $(0.50-x)$. The unknown $q_3$ and $k$ appear on both sides and cancel: $k\frac{q_1{q}_3}{x^2}=k\frac{q_2{q}_3}{(0.50-x{)}^2}\Rightarrow \frac{9\times 10^{-6}}{x^2}=\frac{4\times 10^{-6}}{(0.50-x{)}^2}.$

Step 3 — solve. Cross-multiply and take the square root (both distances are positive, so take the $+$ root): $\frac{9}{x^2}=\frac{4}{(0.50-x{)}^2}\Rightarrow \frac{3}{x}=\frac{2}{0.50-x}\Rightarrow 3(0.50-x)=2x.$ $1.5-3x=2x\Rightarrow 1.5=5x\Rightarrow x=0.30\text{m}.$

Answer: at $x=0.30\text{m}$ — closer to the smaller charge (matching the figure), so the weaker source's shorter distance compensates for its lower charge. The result is independent of the sign and size of $q_3$.

Step 1 — identify the three pushes. Take the corner $P$ at the bottom-left. The other three charges sit at: the adjacent corner along the bottom (distance $a$), the adjacent corner up the side (distance $a$), and the diagonal corner (distance $a\sqrt{2}$). All repel $P$, pushing it away from each source — i.e. down-left overall (the three coloured arrows in the figure).

Step 2 — the two adjacent forces. $F_a=k\frac{q^2}{a^2}=(8.99\times 10^9)\frac{(2\times 10^{-6}{)}^2}{(0.10{)}^2}=(8.99\times 10^9)\frac{4\times 10^{-12}}{0.01}=3.596\text{N}.$ One pushes $P$ straight left (from the bottom-right neighbour), the other straight down (from the top-left neighbour). Put $P$ at the origin with the square in the first quadrant; then these two forces point along $-x$ and $-y$, each of size $3.596\text{N}$.

Step 3 — the diagonal force. Distance $a\sqrt{2}=0.10\sqrt{2}=0.1414\text{m}$, so $r^2=2a^2=0.02$: $F_d=k\frac{q^2}{2a^2}=\frac{F_a}{2}=1.798\text{N}.$ It points from the far corner toward $P$, i.e. along the diagonal into the third quadrant: components $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$ times its size, giving $F_{d,x}=F_{d,y}=-1.798/\sqrt{2}=-1.271$ each.

Step 4 — add components. $F_x=-3.596-1.271=-4.867\text{N},F_y=-3.596-1.271=-4.867\text{N}.$ The two adjacent forces sit along the axes; the diagonal one splits evenly, reinforcing both — so by symmetry $F_x=F_y$, pointing along the outward diagonal.

Step 5 — magnitude. $F_{\text{net}}=\sqrt{F_x^2+F_y^2}=\sqrt{2}(4.867)=6.88\text{N}.$

Answer: $F_{\text{net}}\approx 6.88\text{N}$, directed outward along the diagonal, away from the square's centre (at $45^{\circ }$ below-left of $P$ in our setup).

Compact form to remember: $F_{\text{net}}=\frac{k{q}^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)$. Check: $3.596(1.9142)=6.88$. ✓

Step 1 — what stays, what changes. The ratio of electric force to gravitational force is $\frac{F_e}{F_g}=\frac{k{e}^2/r^2}{G{m}^2/r^2}=\frac{k{e}^2}{G{m}^2},$ where $G=6.67\times 10^{-11}{\text{N⋅m}}^2/{\text{kg}}^2$ is the gravitational constant. Notice the distance $r$ cancels (both forces share the $1/r^2$ shape). Charge magnitude $e$ is identical for electron and proton, and $k,G$ are constants. Only the mass $m$ changes. So the ratio scales as $1/m^2$.

Step 2 — predict. A proton is heavier than an electron by $\frac{m_p}{m_e}=\frac{1.67\times 10^{-27}}{9.11\times 10^{-31}}\approx 1833.$ Since the ratio $\propto 1/m^2$, replacing electrons with protons shrinks it by a factor $1833^2\approx 3.36\times 10^6$: $\frac{F_e}{F_g}{|}_{\text{proton}}\approx \frac{4.2\times 10^{42}}{3.36\times 10^6}\approx 1.25\times 10^{36}.$

Step 3 — verify directly. $\frac{k{e}^2}{G{m}_p^2}=\frac{(8.99\times 10^9)(1.6\times 10^{-19}{)}^2}{(6.67\times 10^{-11})(1.67\times 10^{-27}{)}^2}\approx 1.24\times 10^{36}.✓$ The quick prediction ($1.25\times 10^{36}$) and the full calculation ($1.24\times 10^{36}$) agree.

Answer: about $1.2\times 10^{36}$ — still astronomically large, but roughly $10^6$ times smaller than for electrons, purely because protons are heavier. Distance never entered, because both forces share the $1/r^2$ shape.

Zero net force on the middle of three equal charges? ::: Yes, by symmetry the two equal opposite pushes cancel. Null point between $+9\mu \text{C}$ and $+4\mu \text{C}$ (0.50 m apart)? ::: At 0.30 m from the $9\mu \text{C}$ charge (closer to the smaller one). Net force at a corner of a square of equal charges? ::: $\frac{k{q}^2}{a^2}(\sqrt{2}+\frac{1}{2})$, along the outward diagonal. Why does $r$ never appear in the $F_e/F_g$ ratio? ::: Both forces $\propto 1/r^2$, so distance divides out. Proton-proton electric/gravity ratio? ::: About $1.2\times 10^{36}$, i.e. $(m_p/m_e{)}^2\approx 10^6$ times smaller than the electron value.