Potential of point charge, potential from field and vice versa: Exercises

Level 1 — Recognition

Goal: recognise which relationship applies and read a value off a formula.

L1.1 A point charge $q = + 5 μ C$ sits at the origin. Write the potential $V (r)$ as a function of distance $r$ , then evaluate it at $r = 2 m$ .

Recall Solution L1.1

WHAT we use: potential of a point charge, $V (r) = \frac{k q}{r}$ . WHY: a single point charge — the formula from the parent note applies directly, no integration needed. $V (r) = \frac{( 9 \times 1 0 ^{9} ) ( 5 \times 1 0 ^{- 6} )}{r} = \frac{4.5 \times 1 0 ^{4}}{r} V .$ At $r = 2 m$ : $V = \frac{4.5 \times 1 0 ^{4}}{2} = 2.25 \times 1 0^{4} V = 22500 V .$

L1.2 For a point charge, how does $V$ fall off with distance, and how does the field magnitude $E$ fall off? Which drops faster?

Recall Solution L1.2

$V \propto \frac{1}{r}$ (one power of $r$ on the bottom); $E \propto \frac{1}{r ^{2}}$ (two powers). Because the exponent is bigger for $E$ , the field drops faster than the potential. Double the distance: $V$ halves, but $E$ falls to a quarter.

Level 2 — Application

Goal: substitute correctly, keeping units and signs honest.

L2.1 A charge $q = - 4 μ C$ produces a potential $V = - 1.2 \times 1 0^{4} V$ at some point. How far is that point from the charge?

Recall Solution L2.1

WHAT: invert $V = \frac{k q}{r}$ to solve for $r = \frac{k q}{V}$ . WHY: we know $V$ and $q$ ; distance is the only unknown. Carry the negative signs — both $q$ and $V$ are negative, so $r$ comes out positive (as a distance must). $r = \frac{k q}{V} = \frac{( 9 \times 1 0 ^{9} ) ( - 4 \times 1 0 ^{- 6} )}{- 1.2 \times 1 0 ^{4}} = \frac{- 3.6 \times 1 0 ^{4}}{- 1.2 \times 1 0 ^{4}} = 3 m .$

L2.2 The potential in a region is $V (x) = 5 x$ volts ( $x$ in metres). Find the $x$ -component of the field, $E_{x}$ .

Recall Solution L2.2

WHAT: $E_{x} = - \frac{d V}{d x}$ . WHY: field is minus the slope of the potential (parent §3). $E_{x} = - \frac{d}{d x} (5 x) = - 5 V/m .$ Constant field, pointing in $- x$ (because $V$ rises with $x$ , and field points downhill).

L2.3 A uniform field $E = 300 V/m$ points along $+ x$ . Find the potential difference $V_{A} - V_{B}$ between $x_{A} = 2 m$ and $x_{B} = 5 m$ .

Recall Solution L2.3

WHAT: $V_{A} - V_{B} = - \int_{B}^{A} E d x = - E (x_{A} - x_{B})$ for a uniform field. WHY: the field is constant, so the integral is just $E \times (displacement)$ ; the minus sign is the definition of potential. $V_{A} - V_{B} = - 300 (2 - 5) = - 300 \times (- 3) = 900 V .$ $A$ (lower $x$ ) is at higher potential — moving against the field raises $V$ .

Level 3 — Analysis

Goal: combine several charges or several ideas; symmetry and signs get subtle.

L3.1 Two charges lie on the $x$ -axis: $q_{1} = + 3 μ C$ at $x = 0$ and $q_{2} = + 3 μ C$ at $x = 6 m$ . Find (i) the potential $V$ at the midpoint $x = 3 m$ , and (ii) the field $E$ there.

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

Recall Solution L3.1

Look at the figure: both charges are equal and positive; the midpoint (yellow dot) is $3 m$ from each. (i) Potential — add scalars: $V = k \frac{q _{1}}{r _{1}} + k \frac{q _{2}}{r _{2}} = 9 \times 1 0^{9} (\frac{3 \times 1 0 ^{- 6}}{3} + \frac{3 \times 1 0 ^{- 6}}{3})$ $= 9 \times 1 0^{9} \times 2 \times 1 0^{- 6} = 1.8 \times 1 0^{4} V .$ (ii) Field — add vectors: the field from $q_{1}$ points $+ x$ (red arrow, pushing away from left charge), the field from $q_{2}$ points $- x$ (equal magnitude). They cancel: $E = 0.$ The punchline: $E = 0$ but $V \neq = 0$ at the same point. Field cancellation is about directions; potential is a plain sum of numbers with no direction to cancel.

L3.2 Now make them opposite: $q_{1} = + 3 μ C$ at $x = 0$ , $q_{2} = - 3 μ C$ at $x = 6 m$ . Find $V$ and $∣ E ∣$ at the midpoint $x = 3 m$ .

Recall Solution L3.2

Potential: $V = 9 \times 1 0^{9} (\frac{3 \times 1 0 ^{- 6}}{3} + \frac{- 3 \times 1 0 ^{- 6}}{3}) = 0 V .$ The two signed contributions cancel exactly. Field: each charge's field at the midpoint points in $+ x$ (the $+ q$ pushes right, the $- q$ pulls right), so they add: $E_{1} = E_{2} = \frac{k ∣ q ∣}{r ^{2}} = \frac{9 \times 1 0 ^{9} \times 3 \times 1 0 ^{- 6}}{3 ^{2}} = 3000 V/m,$ $∣ E ∣ = E_{1} + E_{2} = 6000 V/m .$ The punchline: here $V = 0$ but $E \neq = 0$ — the exact opposite of L3.1. This is the parent note's headline mistake made concrete.

Level 4 — Synthesis

Goal: build a new result by chaining the field↔potential machinery.

L4.1 A potential field is $V (x, y) = 4 x^{2} - 3 y$ (volts, metres). Find the full vector field $E$ , then evaluate it at the point $(1, 2)$ . Give the magnitude too.

Recall Solution L4.1

WHAT: $E = - \nabla V = - (\frac{\partial V}{\partial x} \overset{x}{^} + \frac{\partial V}{\partial y} \overset{y}{^})$ . WHY: $V$ depends on two coordinates, so we need the gradient, not a single derivative. Each partial derivative asks "how steeply does $V$ change if I move purely along this axis, holding the other fixed?"

$\frac{\partial V}{\partial x} = 8 x$ (treat $y$ as constant, so $- 3 y$ vanishes).
$\frac{\partial V}{\partial y} = - 3$ (treat $x$ as constant, so $4 x^{2}$ vanishes). $E = - (8 x) \overset{x}{^} - (- 3) \overset{y}{^} = - 8 x \overset{x}{^} + 3 \overset{y}{^} .$ At $(1, 2)$ : $E = - 8 \overset{x}{^} + 3 \overset{y}{^} V/m .$ Magnitude: $∣ E ∣ = (- 8)^{2} + 3^{2} = 64 + 9 = 73 \approx 8.54 V/m .$

L4.2 Consistency check. Start from the point-charge potential $V (r) = \frac{k q}{r}$ , compute $E_{r} = - \frac{d V}{d r}$ , and confirm you recover Coulomb's field. Then, going the other way, integrate that field from $\infty$ to $r$ and confirm you get $V$ back.

Recall Solution L4.2

Forward (potential → field): $E_{r} = - \frac{d}{d r} (\frac{k q}{r}) = - k q \cdot (- \frac{1}{r ^{2}}) = \frac{k q}{r ^{2}} . ✓$ This is exactly the Coulomb field. Back (field → potential): $V (r) = - \int_{\infty}^{r} \frac{k q}{r ^{'2}} d r^{'} = - k q [- \frac{1}{r ^{'}}]_{\infty}^{r} = - k q (- \frac{1}{r} + 0) = \frac{k q}{r} . ✓$ The point: derivative and integral are inverse operations, so the $V \leftrightarrow E$ bridge is perfectly two-way (this is why the field must be conservative — see Conservative Fields and Curl).

Level 5 — Mastery

Goal: full multi-step reasoning, energy, and a limiting case.

L5.1 Three charges sit at the corners of a square of side $a = 1 m$ : $+ 2 μ C$ , $- 2 μ C$ , and $+ 2 μ C$ (going around three corners). Find the potential at the empty fourth corner.

Recall Solution L5.1

Look at the figure: label the empty corner $P$ . Two charges are on adjacent corners (distance $a$ from $P$ ); one charge is on the diagonal corner (distance $a 2$ from $P$ ). From the figure the two adjacent charges are $+ 2 μ C$ and $+ 2 μ C$ (each at distance $a = 1$ ), and the diagonal charge is $- 2 μ C$ (at distance $a 2 = 2$ ). Add scalars, each with its own distance: $V = k (\frac{+ 2 \times 1 0 ^{- 6}}{1} + \frac{+ 2 \times 1 0 ^{- 6}}{1} + \frac{- 2 \times 1 0 ^{- 6}}{2}) .$ $V = 9 \times 1 0^{9} \times 1 0^{- 6} (2 + 2 - \frac{2}{2}) = 9 \times 1 0^{3} (4 - 2) .$ $V = 9000 \times (4 - 1.4142) = 9000 \times 2.5858 \approx 2.327 \times 1 0^{4} V .$ Why the diagonal distance is $a 2$ : the diagonal of a square of side $a$ is $a^{2} + a^{2} = a 2$ (Pythagoras).

L5.2 How much work must an external agent do to bring a charge $Q = + 1 μ C$ from infinity to the corner $P$ of L5.1 (no change in kinetic energy)?

Recall Solution L5.2

WHAT: work $= Q \times V$ where $V$ is the potential at the destination. WHY: by definition (parent §1) $V$ is work per unit charge, so multiply by the charge actually moved. This connects to Potential Energy of Charge System via $U = Q V$ . $W_{ext} = Q V = (1 \times 1 0^{- 6}) (2.327 \times 1 0^{4}) \approx 2.33 \times 1 0^{- 2} J .$ Positive work: we must push the positive $Q$ toward a region of positive potential.

L5.3 (limiting case) Take the L3.2 dipole ( $+ q$ and $- q$ separated by $d$ ) and look at a point on the perpendicular bisector — the line of symmetry midway between them. Show that $V = 0$ everywhere on that line, and check this is consistent as the point goes to infinity.

Recall Solution L5.3

Geometry: any point on the perpendicular bisector is equidistant from both charges — call that distance $s$ . Both charges are the same distance away. Potential: $V = k \frac{+ q}{s} + k \frac{- q}{s} = \frac{k q}{s} - \frac{k q}{s} = 0.$ The two equal-distance contributions cancel for every $s$ , so the whole bisector is a zero-potential (equipotential) plane. Limiting check: as the point recedes ( $s \to \infty$ ), each term $\to 0$ individually, so $V \to 0$ — consistent. The bisector's $V = 0$ is not because the point is far, but because of exact cancellation; the limit just confirms nothing pathological happens far away. Degenerate case: if instead the two charges were equal ( $+ q, + q$ ), the bisector would give $V = \frac{2 k q}{s} \neq = 0$ — cancellation depends entirely on the opposite signs.

Recall wrap-up

Recall One-line answers (cover them)

$V$ at $2 m$ from $+ 5 μ C$ ? ::: $2.25 \times 1 0^{4} V$ .
Midpoint of two equal $+$ charges: $E$ ? $V$ ? ::: $E = 0$ , $V \neq = 0$ .
Midpoint of $+ q, - q$ : $V$ ? $E$ ? ::: $V = 0$ , $E \neq = 0$ .
$E$ from $V = 4 x^{2} - 3 y$ ? ::: $- 8 x \overset{x}{^} + 3 \overset{y}{^}$ .
Work to bring $Q$ to potential $V$ ? ::: $W = Q V$ .

Connections

Coulomb's Law — the $1/ r^{2}$ field behind every $E$ here.
Electric Field of Point Charge — recovered in L4.2.
Potential Energy of Charge System — $U = Q V$ used in L5.2.
Equipotential Surfaces — the $V = 0$ plane of L5.3.
Conservative Fields and Curl — why the L4.2 bridge is two-way.
Gradient Operator — the machinery of L4.1.