Define electric field and electric potential: Worked examples

This page is the practice range for the parent topic. We march through every kind of case the topic can throw at you — positive and negative sources, points on either side, the special "cancel-out" spots, the limits at zero and infinity, a real-world battery problem, and one exam-style twist.

Before touching numbers, let us define the two tools once more, in plain words, so no symbol appears unearned.

The one trap that recurs: $V$ adds like ordinary numbers (it is a scalar), but $E$ adds like arrows (it is a vector — you must track direction). Every example below is really a lesson in which kind of adding you owe.

The scenario matrix

#	Case class	What is tricky	Example
A	Single positive source	direction of $E$ points away	Ex 1
B	Single negative source	$V < 0$ ; $E$ points toward	Ex 2
C	Two charges, point between them	vector add of $E$ ; scalar add of $V$	Ex 3
D	Two equal-opposite charges — the " $V = 0$ , $E \neq = 0$ " spot	the classic dipole midpoint	Ex 4
E	Two equal like charges — the " $E = 0$ , $V \neq = 0$ " spot	field cancels, potential doesn't	Ex 5
F	Limiting behaviour ( $r \to \infty$ , $r \to 0$ )	what blows up, what dies	Ex 6
G	Real-world word problem (battery / charger)	energy from $Δ V$	Ex 7
H	Exam twist: given $V (r)$ , get $E$ by slope	use $E = - \frac{d V}{d r}$	Ex 8

We now walk each cell.

Worked Examples

Worked example Ex 3 — Cell C · two charges, point between them

Charge $Q_{1} = + 5 μ C$ at $x = 0$ , charge $Q_{2} = + 5 μ C$ at $x = 0.4 m$ . Find $E$ and $V$ at the midpoint $x = 0.2 m$ .

Forecast: both charges are equal and positive — will the fields help each other or fight?

Field from $Q_{1}$ at distance $0.2 m$ : $E_{1} = \frac{( 8.99 \times 1 0 ^{9} ) ( 5 \times 1 0 ^{- 6} )}{( 0.2 ) ^{2}} = 1.12375 \times 1 0^{6} N/C$ , pointing right (away from $Q_{1}$ ). Why this step? Compute each source's field separately — fields superpose (add as vectors).
Field from $Q_{2}$ at distance $0.2 m$ : same magnitude $1.12375 \times 1 0^{6} N/C$ , but pointing left (away from $Q_{2}$ , which is on the right). Why this step? At the midpoint the two point in opposite directions — look at the red arrows in the figure.
Net field: $E_{n e t} = E_{1} - E_{2} = 0 N/C$ . Why this step? Equal magnitudes, opposite directions ⟹ they cancel. This is a vector cancellation.
Potential (scalar add): $V = \frac{k Q _{1}}{0.2} + \frac{k Q _{2}}{0.2} = 2 \times \frac{( 8.99 \times 1 0 ^{9} ) ( 5 \times 1 0 ^{- 6} )}{0.2} = 4.495 \times 1 0^{5} V$ . Why this step? Potentials add as plain numbers — no direction, no cancellation, even though the field cancels.

Verify: $E_{n e t} = 0$ but $V \neq = 0$ — this is exactly the "field zero, potential nonzero" phenomenon promised in the parent's mistake box. ✔

Worked example Ex 4 — Cell D · dipole midpoint (

$V = 0$ , $E \neq = 0$ ) $Q_{1} = + 5 μ C$ at $x = 0$ ; $Q_{2} = - 5 μ C$ at $x = 0.4 m$ . Find $E$ and $V$ at the midpoint $x = 0.2 m$ .

Forecast: opposite charges now — do the fields cancel or reinforce? What about $V$ ?

Field from $Q_{1}$ (positive): magnitude $1.12375 \times 1 0^{6} N/C$ , points right (away from +).
Field from $Q_{2}$ (negative): magnitude $1.12375 \times 1 0^{6} N/C$ , points right (toward the − on the right). Why this step? Both arrows now point the same way — see the two red arrows in the figure both aiming right.
Net field: $E_{n e t} = 1.12375 \times 1 0^{6} + 1.12375 \times 1 0^{6} = 2.2475 \times 1 0^{6} N/C$ , pointing right. Why this step? Same direction ⟹ magnitudes add.
Potential: $V = \frac{k ( + 5 μ C )}{0.2} + \frac{k ( - 5 μ C )}{0.2} = 0 V$ . Why this step? Equal-and-opposite scalars sum to zero — this is the " $V = 0$ , $E \neq = 0$ " spot.

Verify: $V = 0$ exactly, $E$ maximal ✔. Note this is the mirror image of Ex 3 — swapping like/opposite charges swaps which quantity cancels.

Worked example Ex 6 — Cell F · limiting behaviour (

$r \to \infty$ and $r \to 0$ ) Take $Q = + 1 μ C$ . Evaluate $E$ and $V$ at (a) $r = 100 m$ and (b) $r = 0.001 m$ , and describe the trend.

Forecast: which quantity blows up faster as you approach the charge — $E$ or $V$ ?

Far away, $r = 100 m$ : $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 1 0 ^{- 6} )}{10 0 ^{2}} = 0.899 N/C$ ; $V = \frac{( 8.99 \times 1 0 ^{9} ) ( 1 0 ^{- 6} )}{100} = 89.9 V$ . Why this step? Both shrink toward $0$ as $r \to \infty$ — infinity is our zero-of-potential reference.
Very close, $r = 0.001 m$ : $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 1 0 ^{- 6} )}{( 0.001 ) ^{2}} = 8.99 \times 1 0^{9} N/C$ ; $V = \frac{( 8.99 \times 1 0 ^{9} ) ( 1 0 ^{- 6} )}{0.001} = 8.99 \times 1 0^{6} V$ . Why this step? As $r \to 0$ , $E \propto 1/ r^{2}$ explodes faster than $V \propto 1/ r$ . Both diverge (a point charge is an idealisation).

Verify: ratio $E / V = 1/ r$ : at $r = 100$ , $0.899/89.9 = 0.01 = 1/100$ ✔; at $r = 0.001$ , $8.99 \times 1 0^{9} /8.99 \times 1 0^{6} = 1000 = 1/0.001$ ✔. The relation $E = V / r$ holds for a single point charge.

Worked example Ex 8 — Cell H · exam twist, get field from a given potential

Suppose along a line the potential is measured to be $V (r) = \frac{6}{r}$ (in volts, $r$ in metres). Find the field magnitude $E$ at $r = 2 m$ .

Forecast: which tool undoes a "value" to give a "slope"? (The derivative.)

Why a derivative? Field is the steepness of the potential hill: $E = - \frac{d V}{d r}$ . We are handed the hill's height $V (r)$ and asked for its slope — that is exactly what a derivative answers.
Differentiate: $\frac{d V}{d r} = \frac{d}{d r} (6 r^{- 1}) = - 6 r^{- 2} = - \frac{6}{r ^{2}}$ . Why this step? Power rule; the negative appears because a $1/ r$ hill slopes downward as $r$ grows.
Apply the sign flip: $E = - \frac{d V}{d r} = + \frac{6}{r ^{2}}$ .
Plug $r = 2$ : $E = \frac{6}{4} = 1.5 N/C$ .

Verify: the given $V = 6/ r$ matches a point charge with $k Q = 6$ , so $E$ should be $\frac{k Q}{r ^{2}} = \frac{6}{r ^{2}} = 1.5 N/C$ at $r = 2$ ✔ — the slope method agrees with the direct formula.

Active Recall

Recall One-line summary of the whole matrix

When do fields cancel but potentials don't? ::: With two equal like charges at a symmetric midpoint — $E = 0$ , $V \neq = 0$ (Ex 3, 5). When do potentials cancel but fields don't? ::: With equal opposite charges (a dipole) at the midpoint — $V = 0$ , $E \neq = 0$ (Ex 4). As $r \to 0$ , which grows faster, $E$ or $V$ ? ::: $E \propto 1/ r^{2}$ grows faster than $V \propto 1/ r$ (Ex 6). How do you get $E$ from a known $V (r)$ ? ::: Take the negative slope: $E = - d V / d r$ (Ex 8). Do you add fields or potentials with signs-and-directions? ::: Fields add as vectors (direction matters); potentials add as scalars (plain signed numbers).

Connections

Define electric field and electric potential (index 1.1.10) — the parent this practice serves.
Coulomb's Law — the force these fields come from.
Electric Potential Energy — $U = q_{0} V$ , the energy the Ex 7 battery delivers.
Voltage and Potential Difference — the $V_{i} - V_{f}$ used in Ex 7.
Equipotential Surfaces and Field Lines — why the dipole midpoint (Ex 4) lies on the $V = 0$ surface.
Gauss's Law — a shortcut to the same fields for symmetric charge.

Two equal like charges at midpoint

E = 0

V \neq = 0

(fields cancel, potentials pile).

Dipole midpoint

V = 0

E \neq = 0

(potentials cancel, fields reinforce).

Field from a given potential

E = - d V / d r

; for

V = 6/ r

E = 6/ r^{2}

Energy from potential difference

W = q (V_{i} - V_{f})

;

2 C

over

5 V

gives

10 J