Define voltage (potential difference) and its units: Worked examples

This page is the "throw everything at it" companion to the voltage topic note. We built the definition $V = W / q$ there; here we drill it against every kind of situation a problem can hand you — positive and negative charges, zero inputs, huge and tiny values, a real-world word problem, and an exam twist.

Before line one, the symbols so nothing is ever unearned:

Recall Every symbol used on this page

$W$ ::: work / energy, measured in joules (J) — the "effort" spent moving charge. $q$ ::: charge, measured in coulombs (C) — "how much electric stuff" you moved. $V$ ::: voltage (potential difference), measured in volts (V) $= J/C$ — energy spent per coulomb. $t$ ::: time, measured in seconds (s) — how long the charge kept moving. $I$ ::: current, measured in amperes (A) $= C/s$ — charge flowing per second, so $I = q / t$ . $P$ ::: power, measured in watts (W) $= J/s$ — energy delivered per second, so $P = W / t = V I$ .

The single equation and its three faces (from the parent note): $V = \frac{W}{q} W = q V q = \frac{W}{V}$

We will also lean on two facts imported from neighbours: current $I = q/t$ (charge per second, rearranges to $q = I t$ ) and power $P = VI$ (energy per second, so $W = P t$ ). The symbol $t$ is just clock time in seconds; every relation that uses it is spelled out where it appears.

Definition The subscript

$V_{A B}$ — reading the order of indices Whenever you see a double-subscript voltage like $V_{A B}$ , read it as "the potential of $A$ measured relative to $B$ ." Formally it is a subtraction: $V_{A B} = V_{A} - V_{B}$ where $V_{A}$ and $V_{B}$ are each point's potential (its "height" measured from whatever we called $0$ ). The order of the letters is a direction: the first letter is the point you're asking about, the second is the reference you compare it to.

If $A$ is higher than $B$ , then $V_{A B} > 0$ (a real uphill).
Swap the order and you flip the sign: $V_{B A} = V_{B} - V_{A} = - V_{A B}$ . Think of it exactly like "how much taller is $A$ than $B$ ?" — reverse the question ("how much taller is $B$ than $A$ ?") and the answer changes sign. This convention is used silently in every example below, so it is worth fixing now.

The scenario matrix

Every voltage problem falls into one of these cells. The examples below are labelled with the cell they cover, so together they leave no gap.

#	Case class	What's tricky about it	Example
A	Plain "find V" — given $W$ and $q$	just divide — the anchor case	Ex 1
B	Find W — given $q$ and $V$	multiply, watch units	Ex 2
C	Find q — given $W$ and $V$	rearrange to $q = W / V$	Ex 3
D	Negative charge / sign of work / $V < 0$	moving a negative charge, and reading a negative $V_{B A}$	Ex 4
E	Zero & degenerate inputs	$V = 0$ (same height), $q = 0$ (nothing moved)	Ex 5
F	Limiting / extreme scale	micro-coulombs and mega-volts — powers of ten	Ex 6
G	Real-world word problem	strip a story down to $W$ , $q$ , $V$	Ex 7
H	Multi-step: current + time + power	chain $I = q / t$ , $W = q V$ , $P = V I$	Ex 8
I	Exam twist: two references / relative voltage	"voltage at a point" is secretly a difference	Ex 9

Keep this picture of "voltage = height of an electric hill" in mind — it is the mental model every example below plugs into.

What the figure below shows: a blue ramp rising from a low landing on the left up to a high landing on the right — this ramp is the electric hill. A green dot marks the low point $B$ ; a red dot marks the high point $A$ . A yellow arrow drags a charge $q$ from $B$ up to $A$ (that trip costs work $W$ ). On the far right a white double-headed arrow spans the vertical gap between the two landings and is labelled "height = voltage $V = W / q$ " — the whole point being that the height of the ramp is what we call the voltage. When Example 1 divides $W$ by $q$ , it is measuring exactly this height; when Example 4 flips the sign for a negative charge, it asks whether the charge slides up or down this same slope; when Example 9 re-zeros the reference, it slides "sea level" up and down without changing the ramp's height.

Figure — Define voltage (potential difference) and its units

Cell A — Plain "find V"

Worked example Example 1 — the anchor case

It takes 20 J of work to move 4 C of charge from point $B$ to point $A$ . Find $V_{A B}$ .

Forecast: Guess before computing — is the answer bigger or smaller than 20? Bigger or smaller than 4? Jot it down.

Identify what we have: $W = 20 J$ , $q = 4 C$ , want $V_{A B}$ (the potential of $A$ relative to $B$ , since we carried charge from $B$ up to $A$ ). Why this step? Matching the givens to the equation's slots tells us which face of $V = W / q$ to use — here energy and charge are known, voltage unknown, so use $V = W / q$ directly. This is literally reading the height of the hill in the figure above.
Divide: $V_{A B} = \frac{W}{q} = \frac{20 J}{4 C} = 5 V$ Why this step? Division answers the literal question "how many joules per coulomb?" — that ratio is the voltage.

Verify: Units: $J / C = V$ ✓. Sanity: $5$ sits between the small carry-cost and the big total, exactly as "per-charge" averaging should. Reverse-check: $q V = 4 \times 5 = 20 J = W$ ✓.

Cell B — Find the energy

Cell C — Find the charge

Cell D — Negative charge, sign of work, and a negative $V$

Voltage doesn't care whether the charge is positive or negative, but the direction energy flows does. This is the same electric hill from the intro figure, now asking whether the charge climbs it or slides down it.

What the figure below shows: two side-by-side copies of the same blue ramp (same voltage, same height), each with a low point $B$ on the left and a high point $A$ on the right. Left panel: a red positive charge (a "+" disc) with a red arrow being dragged uphill from $B$ to $A$ — you must spend energy, so $W = + q V > 0$ . Right panel: a green negative charge (a "−" disc) on the same ramp with a green arrow — the field pulls it uphill for you, so the work you do is $W = q V < 0$ . The message: identical hill, but the sign of the charge decides whether energy is spent or released.

Worked example Example 4 — negative charge (4a) and a negative voltage (4b)

(4a) Between two points the potential difference is $V_{A B} = + 5 V$ ( $A$ is the "high" point). You move a charge of $q = - 2 C$ from $B$ to $A$ . How much work do you do on it? (4b) Now instead move a positive charge $q = + 2 C$ from $A$ down to $B$ . The relevant voltage is now $V_{B A}$ — compute it and the work.

Forecast: In (4a): positive charges resist going uphill, so you spend energy — a negative charge is the opposite, does it need pushing or go on its own? In (4b): guess the sign of $V_{B A}$ before computing.

(4a) Use the definition literally: $W = q V_{A B}$ . Why this step? $V_{A B} = W / q$ is the work per unit charge to move charge from $B$ to $A$ ; multiplying by the actual charge (sign included) gives the actual work.
Substitute with the sign kept: $W = q V_{A B} = (- 2 C) (+ 5 V) = - 10 J$ Why this step? We keep the minus sign because the sign of the charge is physics, not decoration.
Read the sign: $W = - 10 J$ means you did negative work — i.e. the field pushed the negative charge uphill for you, and you'd have to hold it back. The negative charge is attracted toward the high-potential point. Why this step? Negative work is the answer to "who supplied the energy?" — here the field did, not you.
(4b) First find the voltage for the reverse trip using the index rule $V_{B A} = - V_{A B}$ : $V_{B A} = - V_{A B} = - 5 V$ Why this step? Going from $A$ (high) to $B$ (low) is downhill, so its potential difference must be negative — this is precisely the "negative $V$ " case the intro definition flagged. The order of the subscripts flipped, so the sign flipped.
Now the work on the positive charge: $W = q V_{B A} = (+ 2 C) (- 5 V) = - 10 J$ Why this step? A negative voltage means a downhill move, so a positive charge releases energy ( $W < 0$ ): the field does the work, exactly like a rock rolling downhill.

Verify (4a): Flip the charge sign: a $+ 2 C$ charge climbing to $A$ would need $W = (+ 2) (+ 5) = + 10 J$ (you push it uphill). The magnitude matches; only the sign flipped ✓. Verify (4b): $V_{B A} = - V_{A B}$ so $- 5 = - (+ 5)$ ✓, and $W = (2) (- 5) = - 10 J$ ; units $C \times V = J$ ✓. Both "energy released" answers are negative, consistent with downhill motion ✓.

Cell E — Zero and degenerate inputs

Worked example Example 5 — the two "nothing happens" cases

(a) Two points are at the same potential. You move $q = 3 C$ between them. Work? Voltage? (b) A 9 V battery, but you move $q = 0 C$ (no charge at all). Energy delivered?

Forecast: Which quantity goes to zero in each case — and does dividing ever blow up?

(a) Same height ⇒ $V = 0$ . By definition the potential difference is zero: $V_{A B} = V_{A} - V_{B} = 0$ when $V_{A} = V_{B}$ . In the hill picture the two points sit at the same level — the slope is flat. Work: $W = q V = (3 C) (0 V) = 0 J$ . Why this step? No hill means no uphill effort — moving charge sideways on flat ground stores no energy. This is the "no height difference → no flow" line from the parent note, made numeric.
(b) No charge ⇒ $q = 0$ . Energy: $W = q V = (0 C) (9 V) = 0 J$ . Why this step? Voltage is only a readiness to give energy; with zero charge to give it to, nothing is delivered. The 9 V is still "there," but idle.
Degenerate warning: what about $V = W / q$ with $q = 0$ ? That is division by zero — undefined, not zero. You may only compute $V$ from a nonzero charge that actually moved. Why this step? Every formula has a domain; $V = W / q$ requires $q \neq = 0$ . Reach for $W = q V$ (multiplication, always safe) when charge might be zero.

Verify: (a) $W = 0$ : units $C \times V = J$ , and $0$ is dimensionally fine ✓. (b) $W = 0$ ✓. The undefined case correctly refuses to give a number ✓.

Cell F — Limiting / extreme scale

Cell G — Real-world word problem

Worked example Example 7 — the smartphone charge

A phone battery is rated $3.7 V$ and stores $2000 mA\cdoth$ of charge. (a) Convert the charge to coulombs. (b) How much energy does it hold, in joules?

Forecast: $2000 mA\cdoth$ sounds huge — but is it thousands of coulombs, or fewer?

Decode $mA\cdoth$ : it's current × time = charge. Here $m$ means milli $= 1 0^{- 3}$ , so $2000 mA = 2 A$ , and $1 h = 3600 s$ . Why this step? $q = I t$ from $I = q/t$ (rearranged), where $t$ is time in seconds; an amp-hour is literally 1 amp for 1 hour, so it is a charge in disguise.
Compute charge: $q = I t = (2 A) (3600 s) = 7200 C$ Why this step? Multiplying charge-per-second ( $I$ ) by seconds ( $t$ ) gives total charge — the seconds cancel.
Compute energy with the battery voltage: $W = q V = (7200 C) (3.7 V) = 26640 J \approx 2.66 \times 1 0^{4} J$ Why this step? Each coulomb the battery pushes is lifted through $3.7 V$ , so total energy = charge × voltage.

Verify: $q = I t$ : $A \times s = C$ ✓. $W = q V$ : $C \times V = J$ ✓. Cross-check via watt-hours: $26640 J \div 3600 = 7.4 W\cdoth$ , a believable phone rating ✓.

Cell H — Multi-step: current, time, power

Worked example Example 8 — chaining every relation

A $12 V$ car battery drives $3 A$ through a headlight for $2$ minutes. Find (a) the charge moved, (b) the energy delivered, (c) the power — and confirm the two energy routes agree.

Forecast: Two different paths to the energy exist. Will they match to the joule?

Time in seconds: $t = 2 min = 120 s$ . Why this step? SI relations use seconds ( $t$ ), not minutes.
Charge: $q = I t = (3 A) (120 s) = 360 C$ . Why this step? $I = q / t$ rearranges to $q = I t$ ; current × time = total charge.
Energy route 1 (via charge): $W = q V = (360 C) (12 V) = 4320 J$ . Why this step? Charge × voltage = energy, the $W = q V$ face.
Power: $P = V I = (12 V) (3 A) = 36 W$ . Why this step? $P=VI$ is energy per second; it needs only voltage and current.
Energy route 2 (via power): $W = P t = (36 W) (120 s) = 4320 J$ . Why this step? Power × time = energy; this is an independent path, so agreement is a genuine check.

Verify: Both routes give $4320 J$ ✓. Units: route 1 $C\cdotV = J$ ; route 2 $W\cdots = J$ ✓. The chain $P = W / t = q V / t = V (q / t) = V I$ is exactly the parent-note derivation, confirmed numerically.

Cell I — Exam twist: relative voltage / two references

Worked example Example 9 — "the voltage at a node"

In a circuit, point $A$ reads $+ 8 V$ and point $B$ reads $+ 3 V$ , both measured relative to ground (0 V). (a) What is $V_{A B}$ ? (b) You move $q = 2 C$ from $B$ to $A$ — work done? (c) If you re-zero the meter so that $B$ becomes the new $0 V$ reference, what does $A$ read now, and does the work change?

Forecast: Changing the reference changes the individual readings — but does it change the difference?

(a) A single "voltage at a point" is secretly a difference from ground: $V_{A} = 8 V$ means $V_{A} - V_{ground} = 8$ . Apply the index rule $V_{A B} = V_{A} - V_{B}$ : $V_{A B} = V_{A} - V_{B} = 8 V - 3 V = 5 V$ Why this step? Only differences are physical; subtracting the two ground-referenced readings gives the true point-to-point voltage. In the hill picture, ground is just where we painted "sea level."
(b) $W = q V_{A B} = (2 C) (5 V) = 10 J$ . Why this step? Work depends on the difference $V_{A B}$ , not on either absolute reading. Since $V_{A B} > 0$ ( $A$ is uphill of $B$ ) and the charge is positive, this is genuine uphill work — you supply the 10 J.
(c) Re-zeroing at $B$ shifts every reading down by 3 (we subtract 3 from all points so that $B$ lands on 0): $A$ now reads $8 - 3 = 5 V$ , $B$ reads $3 - 3 = 0 V$ . Recompute the difference: $V_{A B} = V_{A} - V_{B} = 5 V - 0 V = 5 V$ so $W = q V_{A B} = (2) (5) = 10 J$ — unchanged. Why this step? This is the whole point: the "hill height" between two spots doesn't care where you set sea level. Choosing a reference shifts every label by the same constant, and that constant cancels in the subtraction $V_{A} - V_{B}$ .

Verify: $V_{A B}$ = 5 V both before ( $8 - 3$ ) and after ( $5 - 0$ ) re-zeroing ✓; work = 10 J both times ✓. The constant shift ( $- 3$ ) cancels: $(V_{A} - 3) - (V_{B} - 3) = V_{A} - V_{B}$ ✓. Matches the parent note's rule "a single point has no voltage until you pick a reference" ✓.

One-glance summary of the matrix

Connections

Electric charge and the coulomb — the $q$ in every example.
Electric current and the ampere — $I = q / t$ , used in Examples 7 and 8.
Power in electric circuits P=VI — the second energy route in Example 8.
Electric field and potential energy — why negative charges "fall uphill" in Example 4.
Ohm's Law V=IR — the natural next step once current enters.
Batteries and EMF — the fixed potential differences in Examples 2, 7, 9.