Worked examples — Potential of point charge, potential from field and vice versa

Two facts we lean on the whole way (proven in the parent):

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The scenario matrix

Every problem this topic can throw is one of these cells. Each is claimed by at least one worked example below.

#	Cell (case class)	What makes it tricky	Example
C1	Single positive charge, get $V$ at a distance	baseline, sign of $V$	Ex 1
C2	Single negative charge, get $V$	$V<0$ — a "valley"	Ex 1
C3	Two charges, scalar superposition (mixed signs)	add signed values, not vectors	Ex 2
C4	Degenerate zero of $\vec{E}$ but $V\ne 0$	field cancels, potential doesn't	Ex 3
C5	Degenerate zero of $V$ but $\vec{E}\ne 0$	potential cancels, field doesn't	Ex 3
C6	$\vec{E}$ from $V(x)$ (1-D slope, sign of $E$)	minus-sign of gradient	Ex 4
C7	$\vec{E}$ from $V(x,y,z)$ (partial derivatives)	vector from scalar	Ex 5
C8	$\Delta V$ from a uniform field	direction of integration, sign	Ex 6
C9	Limiting behaviour $r\to \infty$ and $r\to 0$	how fast $V,E$ die/blow up	Ex 7
C10	Real-world word problem (energy $U=qV$)	translate words → numbers	Ex 8
C11	Exam twist: $V=0$ point on the line of a dipole-like pair	solve for a location	Ex 9

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Ex 1 — one charge, both signs (cells C1, C2)

Forecast: guess the two signs and rough size before reading on. Positive charge → hill top → $V>0$; negative → valley → $V<0$. Same magnitude.

1. Write the formula. $V=\frac{kq}{r}$. Why this step? One point charge, and $V$ needs only the distance $r$, never a direction — that is the whole reason $V$ is a scalar.
2. (a) plug in $q=+4\times 10^{-9}$, $r=2$. $V=\frac{(9\times 10^9)(4\times 10^{-9})}{2}=\frac{36}{2}=18\mathrm{V}.$ Why this step? Pure arithmetic; units $\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}\cdot C/m=\mathrm{N}m/C=J/C=\mathrm{V}$. ✓
3. (b) flip the sign of $q$. $V=-18\mathrm{V}$. Why this step? The sign of the charge rides straight into $V$ — no absolute values, ever.

Verify: $|V_a|=|V_b|$ as expected, and doubling $r$ to $4\mathrm{m}$ would halve each to $\pm 9\mathrm{V}$ because $V\propto 1/r$. Sign flips with charge, magnitude fixed by distance. ✓

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Ex 2 — scalar superposition, mixed signs (cell C3)

Forecast: the point is close to the positive charge ($1\mathrm{m}$) and far from the negative ($3\mathrm{m}$), so guess: net positive, but not huge.

1. Read distances off the picture. From $q_1$: $r_1=|1-0|=1\mathrm{m}$. From $q_2$: $r_2=|1-4|=3\mathrm{m}$. Why this step? $V$ ignores direction — a distance is a plain positive length. Look at the two teal brackets in the figure.
2. Add signed scalar contributions. $V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}=9\times 10^9\left(\frac{2\times 10^{-6}}{1}+\frac{-3\times 10^{-6}}{3}\right).$ Why this step? Superposition for potential is algebra, not trigonometry — signs come from the charges, not from geometry.
3. Compute the bracket, then multiply. $=9\times 10^9(2\times 10^{-6}-1\times 10^{-6})=9\times 10^9\times 10^{-6}=9000\mathrm{V}.$

Verify: result is positive (matches forecast) and finite. Units: $(\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2})(\mathrm{C}{\mathrm{m}}^{-1})=\mathrm{V}$. ✓

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Ex 3 — the two famous degeneracies (cells C4, C5)

Forecast: In one case the field dies but potential is big; in the other the potential dies but field survives. Guess which is which.

Case (A): two like $+q$ (top of figure).

1. Field by symmetry. The left charge pushes right ($+x$), the right charge pushes left ($-x$), equal magnitude ⇒ $\vec{E}=0$ at the origin. Why this step? Field is a vector; equal-and-opposite arrows cancel. See the two red arrows meeting head-on.
2. Potential by scalar sum. Both are positive and equidistant: $V=k\frac{q}{a}+k\frac{q}{a}=\frac{2kq}{a}=\frac{2(9\times 10^9)(10^{-6})}{1}=1.8\times 10^4\mathrm{V}.$ Why this step? Scalars can't cancel by direction — two hills add. So $\vec{E}=0$ yet $V=18000\mathrm{V}\ne 0$. Cell C4.

Case (B): $+q$ left, $-q$ right (bottom of figure).

3. Potential by scalar sum. $V=k\frac{+q}{a}+k\frac{-q}{a}=0.$ Why this step? A hill and an equal valley at equal distance cancel as numbers → $V=0$.
4. Field by vectors. Left $+q$ pushes a test $+$ charge right; right $-q$ pulls it right. Both point $+x$, they add: $E=2\cdot \frac{kq}{a^2}=2\frac{(9\times 10^9)(10^{-6})}{1^2}=1.8\times 10^{4}V/m(\text{along }+x)\ne 0.$ Why this step? Vectors that point the same way reinforce. So $V=0$ yet $\vec{E}\ne 0$. Cell C5.

Verify: the two cases are mirror images — in (A) the vectors cancel and scalars add; in (B) the scalars cancel and vectors add. This is the definitive counter-example to "$V=0\iff E=0$." ✓

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Ex 4 — field from a 1-D potential, watch the sign (cell C6)

Forecast: at $x=2$ the parabola is rising steeply, so the ball should roll toward smaller $x$ — expect $E_x<0$.

1. Differentiate and negate. $E_x=-\frac{dV}{dx}=-(6x-2)$. Why this step? Field is minus the slope of potential — the minus makes the arrow point downhill. Without it energy conservation breaks.
2. Evaluate at $x=2$. $E_x=-(6\cdot 2-2)=-(10)=-10V/m$. Why this step? Just plug in. Negative ⇒ points in $-x$, exactly as forecast (downhill from the rising $V$).

Verify: at the vertex the slope is zero: $6x-2=0\Rightarrow x=\frac{1}{3}$, so $E_x=0$ there — a flat spot has no field, consistent with the slope picture. ✓ Units: $V/m$. ✓

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Ex 5 — vector field from a 3-D potential (cell C7)

Forecast: three partial derivatives → three components. Expect a genuine 3-D arrow.

1. Take each partial derivative (treat the other variables as constants). $\frac{\partial V}{\partial x}=4xy,\frac{\partial V}{\partial y}=2x^2,\frac{\partial V}{\partial z}=-1.$ Why this step? The gradient $\nabla V$ stacks these three slopes into one vector; each one measures how $V$ climbs along that axis.
2. Negate to get the field. $\vec{E}=-\nabla V=(-4xy,-2x^2,+1)$. Why this step? Same minus sign as 1-D — the field points where $V$ decreases fastest.
3. Evaluate at $(1,1,3)$. $\vec{E}=(-4,-2,1)V/m$. Why this step? Plug the coordinates in; note $z$ dropped out because $\partial V/\partial z$ was constant.

Verify: magnitude $|\vec{E}|=\sqrt{(-4{)}^2+(-2{)}^2+1^2}=\sqrt{21}\approx 4.58V/m$. The three components are independent — that's the vector nature $V$ hides but $\vec{E}$ reveals. ✓

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Ex 6 — potential difference from a uniform field (cell C8)

Forecast: field points $+x$ (downhill is $+x$), so lower $x$ = higher ground. $A$ (at $x=1$) sits higher than $B$ (at $x=4$): expect $V_A-V_B>0$.

1. Use $V_A-V_B=-{\int }_B^A\vec{E}\cdot d\vec{l}$. With $\vec{E}$ constant along $x$: $V_A-V_B=-E(x_A-x_B).$ Why this step? Uniform field pulls straight out of the integral; only the endpoint displacement $x_A-x_B$ remains.
2. Plug numbers. $V_A-V_B=-200(1-4)=-200(-3)=600\mathrm{V}$. Why this step? Arithmetic. Positive ⇒ $A$ is at higher potential — matches the forecast (you climb going against the field).

Verify: flip endpoints: $V_B-V_A=-200(4-1)=-600\mathrm{V}$, the negative of our answer — as any well-defined potential difference must be. Units $(V/m)\cdot \mathrm{m}=\mathrm{V}$. ✓

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Ex 7 — limiting behaviour (cell C9)

Forecast: doubling $r$ should halve $V$ but quarter $E$, because $V\propto 1/r$ and $E\propto 1/r^2$.

1. Values at $r=1\mathrm{m}$. $V=\frac{(9\times 10^9)(10^{-6})}{1}=9000\mathrm{V},E=\frac{(9\times 10^9)(10^{-6})}{1^2}=9000V/m.$ Why this step? Baseline to compare against.
2. Values at $r=2\mathrm{m}$. $V=\frac{9000}{2}=4500\mathrm{V}(\times \frac{1}{2}),E=\frac{9000}{2^2}=2250V/m(\times \frac{1}{4}).$ Why this step? Confirms the different power laws — $V$ dies slower than $E$.
3. Limits. As $r\to \infty$: both $\to 0$ (our chosen reference). As $r\to 0$: both blow up ($V\to \infty$, $E\to \infty$) — the idealised point charge is a mathematical singularity, not a real object. Why this step? Every good analysis names its extreme cases so no reader is surprised.

Verify: ratio check $V(1)/V(2)=2$ and $E(1)/E(2)=4$ — exactly the $1/r$ vs $1/r^2$ signatures. ✓

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Ex 8 — real-world word problem, $U=qV$ (cell C10)

Forecast: a positive charge rolls from high $V$ to low $V$ (downhill), so it should gain kinetic energy.

1. Energy is charge times potential: $U=qV$. The change in potential energy is $\Delta U=q(V_{\text{final}}-V_{\text{initial}})=q(20-50)=q(-30).$ Why this step? Potential energy of a charge in a field is $U=qV$; only the change matters for motion.
2. Energy conservation: kinetic gained $=-\Delta U$. $\Delta K=-\Delta U=-q(-30)=30q=30(1.6\times 10^{-19})=4.8\times 10^{-18}\mathrm{J}.$ Why this step? Total energy is conserved, so PE lost becomes KE gained.

Verify: positive $\Delta K$ ⇒ speeds up, matching the "positive charge rolls downhill" forecast. In electron-volts, $\Delta K=q\cdot 30\mathrm{V}=30\mathrm{eV}$ — a clean sanity check, since $1\mathrm{eV}$ is exactly one electronic charge across $1\mathrm{V}$. ✓

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Ex 9 — exam twist: find the point where $V=0$ (cell C11)

Forecast: the positive charge is stronger, so its hill must be "diluted" to zero by being farther from the (weaker) valley — expect the null closer to the negative charge, i.e. $x>3$.

1. Set total potential to zero at position $x$ (with $0<x<6$, so $r_1=x$, $r_2=6-x$): $k\frac{q_1}{x}+k\frac{q_2}{6-x}=0\Rightarrow \frac{2}{x}+\frac{-1}{6-x}=0.$ Why this step? $V=0$ is one scalar equation — no vectors, so we just add signed terms and set the sum to zero (the $k$ and the $10^{-6}$ cancel out).
2. Solve. $\frac{2}{x}=\frac{1}{6-x}\Rightarrow 2(6-x)=x\Rightarrow 12-2x=x\Rightarrow 3x=12\Rightarrow x=4\mathrm{m}$. Why this step? Cross-multiply; one linear equation, one root in range.

Verify: at $x=4$: $V\propto \frac{2}{4}+\frac{-1}{2}=0.5-0.5=0$ ✓. And $x=4>3$, closer to the negative charge — exactly the forecast. Note $\vec{E}\ne 0$ here (Ex 3's lesson: a $V=0$ spot is not a field-free spot). ✓

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Recall

Recall Which cell is which trap?

Field zero but potential big ::: midpoint of two like charges (Ex 3A). Potential zero but field nonzero ::: midpoint of two opposite charges (Ex 3B). Add potentials how? ::: signed scalars, no components (Ex 2). Get $\vec{E}$ from $V$? ::: minus the gradient, $E_x=-dV/dx$ (Ex 4, 5). Double $r$: what happens to $V$ and $E$? ::: $V$ halves, $E$ quarters (Ex 7).

Connections

• Coulomb's Law · Electric Field of Point Charge · Potential Energy of Charge System · Equipotential Surfaces · Conservative Fields and Curl · Gradient Operator