Electric field — definition, field lines, superposition: Worked examples

Before anything, the two symbols we lean on the whole page:

The scenario matrix

Every problem in this topic is one of these cells. The examples below are tagged with the cell they cover.

#	Cell (case class)	What makes it different	Covered by
A	Single positive charge	field points away	Ex 1
B	Single negative charge	field points toward; sign flips $\overset{r}{^}$	Ex 2
C	Two like charges, symmetric point	horizontal parts cancel, vertical add	Ex 3
D	Two unlike (dipole), symmetric point	vertical cancel, horizontal add	Ex 4
E	Limiting behaviour — far away	check both C and D far-field	Ex 3, 4
F	Degenerate / zero-field point	fields exactly cancel to $0$	Ex 5
G	Symmetry gives zero (ring centre)	continuous charge, everything cancels	Ex 6
H	Real-world word problem	translate words → numbers	Ex 7
I	Exam twist — where to put a third charge so it feels no force	solve $E = 0$	Ex 8

Prerequisite tools you'll see used: Coulomb's Law, Superposition Principle, and for the dipole Electric Dipole.

Ex 1 — Cell A: single positive charge

Worked example A charge

$+ 2 nC$ sits at the origin. Find $E$ at the point $3 cm$ to its right on the x-axis. Forecast: Guess before computing — will the field point left or right, and roughly how big (bigger or smaller than $1 0^{4} N/C$ )?

Step 1 — write the knowns in SI units. $q = 2 \times 1 0^{- 9} C$ , $r = 3 \times 1 0^{- 2} m$ . Why this step? $k$ is in metres and coulombs; mixing in nC or cm silently multiplies your answer by billions. Convert first, always.

Step 2 — magnitude from the point-charge formula. $E = \frac{k q}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 9} )}{( 3 \times 1 0 ^{- 2} ) ^{2}}$ Why this step? This is the whole content of Coulomb's Law divided by the test charge — the field of one source.

Step 3 — arithmetic. Numerator $= 17.98$ . Denominator $= 9 \times 1 0^{- 4}$ . $E = \frac{17.98}{9 \times 1 0 ^{- 4}} \approx 1.998 \times 1 0^{4} N/C \approx 2.0 \times 1 0^{4} N/C .$

Step 4 — direction. $q > 0$ , so $E$ points along $\overset{r}{^}$ , i.e. away from the charge — here, to the right (in $+ x$ ). See the red arrow.

Verify: Units: $\frac{( N m ^{2} / C ^{2} ) ( C )}{m ^{2}} = N/C$ ✓. Direction away from $+ q$ ✓, matching the forecast if you guessed "right".

Ex 2 — Cell B: single negative charge

Worked example Replace the charge with

$- 2 nC$ . Find $E$ at the same point. Forecast: Same size number? Same direction? Guess.

Step 1 — magnitude uses $∣ q ∣$ . Magnitude only cares about how much charge: $E = \frac{k ∣ q ∣}{r ^{2}} = 2.0 \times 1 0^{4} N/C$ — identical to Ex 1. Why this step? Distance and amount of charge are unchanged; only the sign changed, and sign is a direction fact, not a size fact.

Step 2 — the sign flips direction. In $E = k q \overset{r}{^} / r^{2}$ , a negative $q$ makes $E$ point opposite to $\overset{r}{^}$ . Since $\overset{r}{^}$ points from the charge outward (rightward), $E$ now points left, toward the charge. Why this step? This is exactly the sign-rule warned about in the parent's mistake box — the sign of $q$ , not the direction of $\overset{r}{^}$ , encodes attract/repel.

Verify: A tiny positive test charge placed there would be pulled toward the negative source — leftward — and $F = q_{0} E$ points along $E$ . Consistent ✓.

Ex 3 — Cell C + E: two equal $+ q$ , symmetric point, and the far limit

Worked example Two charges

$+ q = + 3 nC$ sit at $(\pm a, 0)$ with $a = 4 cm$ . Find $E$ at $(0, y)$ with $y = 3 cm$ . Forecast: Will the answer point up, down, sideways, or be zero?

Step 1 — distance from each charge to the point. By Pythagoras, $r = a^{2} + y^{2} = 4^{2} + 3^{2} cm = 5 cm = 0.05 m .$ Why this step? Each charge sits at a corner of a right triangle with the field point; the hypotenuse is the separation the $1/ r^{2}$ needs.

Step 2 — magnitude from one charge. $E_{1} = \frac{k q}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{( 0.05 ) ^{2}} = \frac{26.97}{0.0025} \approx 1.079 \times 1 0^{4} N/C .$

Step 3 — kill the horizontal parts by symmetry. The left charge pushes up-and-right, the right charge pushes up-and-left. Their $x$ -components are equal and opposite → cancel. Only the upward ( $+ y$ ) parts survive. Why this step? Mirror symmetry about the y-axis: reflecting swaps the two charges but leaves the point fixed, so $E_{x}$ must equal its own negative, i.e. $E_{x} = 0$ .

Step 4 — vertical fraction. The upward slice of each field is $E_{1} cos θ$ , where $cos θ = y / r = 3/5 = 0.6$ . Two charges: $E_{y} = 2 E_{1} cos θ = \frac{2 k q y}{( a ^{2} + y ^{2} ) ^{3/2}} = 2 (1.079 \times 1 0^{4}) (0.6) \approx 1.295 \times 1 0^{4} N/C (upward) .$ Why the $3/2$ power? One power of $r^{2}$ from Coulomb, half a power more from projecting onto $y$ via $cos θ = y / r$ .

Step 5 — the far limit ( $y ≫ a$ ). For $y ≫ a$ , $(a^{2} + y^{2})^{3/2} \to y^{3}$ , so $E_{y} \to \frac{2 k q}{y ^{2}} .$ Why check this? Far away the pair looks like a single lump of charge $2 q$ — and indeed we recover a point-charge $1/ y^{2}$ law. Sanity confirmed.

Verify: Points straight up ✓ (no sideways, matching the symmetry forecast). Number checked in VERIFY.

Ex 4 — Cell D + E: dipole, symmetric point, and its far limit

Worked example Now flip the right charge:

$+ q$ at $(- a, 0)$ is replaced so we have $+ 3 nC$ at $(+ a, 0)$ and $- 3 nC$ at $(- a, 0)$ , same $a = 4 cm$ , $y = 3 cm$ . Find $E$ . Forecast: Which parts cancel this time — the ups or the sideways?

Step 1 — same $r$ , same $E_{1}$ . Geometry didn't change: $r = 0.05 m$ , $E_{1} \approx 1.079 \times 1 0^{4} N/C$ .

Step 2 — which components cancel now. The $+ q$ (right) field points away from it — up-and-left at our point. The $- q$ (left) field points toward it — down-and-left. The vertical parts are now opposite → cancel; the horizontal ( $- x$ ) parts point the same way → add. Why this step? Swapping one charge's sign flips that charge's whole field vector, converting the cancelling pair into the adding pair. This is the heart of what a Electric Dipole is.

Step 3 — horizontal fraction. Here the relevant slice uses $cos ϕ = a / r = 4/5 = 0.8$ : $E = 2 E_{1} cos ϕ = \frac{2 k q a}{( a ^{2} + y ^{2} ) ^{3/2}} = 2 (1.079 \times 1 0^{4}) (0.8) \approx 1.726 \times 1 0^{4} N/C (in - x) .$

Step 4 — far limit ( $y ≫ a$ ). $(a^{2} + y^{2})^{3/2} \to y^{3}$ , so with dipole moment $p = 2 q a$ : $E \to \frac{2 k q a}{y ^{3}} = \frac{k p}{y ^{3}} .$ Why steeper ( $1/ y^{3}$ ) than a point charge? Far off, the $+$ and $-$ almost cancel; only their tiny offset $2 a$ survives, and that extra "smallness" costs one more power of distance.

Verify: Points sideways (in $- x$ ) ✓, opposite to the like-charge case's vertical answer — the two symmetries are complementary. Number in VERIFY.

Ex 5 — Cell F: the exact zero-field point

Worked example Two charges:

$+ 4 nC$ at $x = 0$ and $+ 1 nC$ at $x = 0.30 m$ . On the line joining them, where is $E = 0$ ? Forecast: Nearer the big charge or the small one? Guess before solving.

Step 1 — where could it cancel? Between two positive charges, the two fields point in opposite directions only in the gap between them. Outside, both point the same way and can't cancel. Why this step? Cancellation needs opposition; only the interior region offers it.

Step 2 — set magnitudes equal. Let the null point be at distance $x$ from the $+ 4 nC$ charge, so $0.30 - x$ from the $+ 1 nC$ : $\frac{k ( 4 \times 1 0 ^{- 9} )}{x ^{2}} = \frac{k ( 1 \times 1 0 ^{- 9} )}{( 0.30 - x ) ^{2}} .$ Why this step? $E = 0$ means the two magnitudes match (directions already oppose). $k$ and $1 0^{- 9}$ cancel.

Step 3 — solve. $4 (0.30 - x)^{2} = x^{2} \Rightarrow 2 (0.30 - x) = x$ (take positive root): $0.60 - 2 x = x \Rightarrow x = 0.20 m .$ So the null point is $0.20 m$ from the big charge, $0.10 m$ from the small one.

Verify: It sits closer to the smaller charge ✓ (a weaker charge needs you nearer to match the stronger one). Ratio of distances $0.20/0.10 = 2 = 4/1$ ✓ — exactly $q_{1} / q_{2}$ .

Ex 6 — Cell G: symmetry ⇒ zero (ring centre)

Worked example A thin ring of radius

$R$ carries total charge $Q$ spread uniformly. Show $E$ at the centre is zero, using Continuous Charge Distributions. Forecast: With charge all around you, is the centre field large or zero?

Step 1 — chop the ring into tiny pieces $d q$ . Each little arc is nearly a point charge; it makes $d E = \frac{k d q}{R ^{2}}$ pointing from that piece toward the centre. Why this step? We can't apply the point-charge formula to a smear directly, so we superpose over infinitesimal points and integrate.

Step 2 — pair opposite pieces. For every piece at angle $θ$ there is an identical piece at $θ + 18 0^{\circ}$ directly across. Their two $d E$ are equal in size and exactly opposite in direction.

Step 3 — the sum collapses. Every pair cancels, so $E_{centre} = \oint d E = 0 .$ Why this step? The integral of perfectly balanced opposite vectors is zero — no arithmetic needed once the symmetry is seen.

Verify: Independent of $Q$ and $R$ — as long as the ring is uniform, the centre is always a null. This is the same reasoning Gauss's Law exploits: symmetry does the work.

Ex 7 — Cell H: real-world word problem

Worked example A tiny paint droplet carries charge

$- 5.0 nC$ and mass $2.0 \times 1 0^{- 8} kg$ . It hovers motionless in air. What electric field (size and direction) holds it up? ( $g = 9.8 m/ s^{2}$ .) Forecast: Field points up or down, given the charge is negative?

Step 1 — force balance. Hovering means net force zero: the electric force up must equal gravity down. $∣ q ∣ E = m g .$ Why this step? "Motionless" is the physics translation of $\sum F = 0$ ; this is where energy ideas aren't needed — plain Newton.

Step 2 — solve for $E$ . $E = \frac{m g}{∣ q ∣} = \frac{( 2.0 \times 1 0 ^{- 8} ) ( 9.8 )}{5.0 \times 1 0 ^{- 9}} = \frac{1.96 \times 1 0 ^{- 7}}{5.0 \times 1 0 ^{- 9}} = 39.2 N/C .$

Step 3 — direction. The force on the droplet must be upward. Force on a charge is $F = q E$ ; with $q < 0$ , $F$ is opposite to $E$ . To get force up, $E$ must point down. Why this step? Same sign logic as Ex 2 — never read direction off the force alone when the charge is negative.

Verify: Units $\frac{kg \cdot m/ s ^{2}}{C} = N/C$ ✓. $E$ downward, force $q E$ upward, balances gravity ✓.

Ex 8 — Cell I: exam twist (where does a third charge feel no force?)

Worked example Charges

$+ q$ at $x = 0$ and $+ 9 q$ at $x = 1.0 m$ . A third charge $Q$ is to be placed on the line so it feels no force. Where — and does the answer depend on the sign or size of $Q$ ? Forecast: Nearer the $+ q$ or the $+ 9 q$ ?

Step 1 — "no force" means " $E = 0$ at that spot". Since $F = Q E$ , the force vanishes exactly where the net field of the other two vanishes — independent of $Q$ . Why this step? This is the key twist: it reduces a force problem to the Ex-5 field-null problem. $Q$ 's own field never acts on itself (parent mistake box).

Step 2 — set field magnitudes equal (interior of the gap). Let it sit at distance $x$ from the $+ q$ : $\frac{k q}{x ^{2}} = \frac{k ( 9 q )}{( 1 - x ) ^{2}} \Rightarrow (1 - x)^{2} = 9 x^{2} \Rightarrow 1 - x = 3 x .$

Step 3 — solve. $1 = 4 x \Rightarrow x = 0.25 m$ from the $+ q$ charge.

Verify: Closer to the smaller charge $+ q$ ✓. Distance ratio $0.25 : 0.75 = 1 : 3 = 1 : 9$ ✓. Answer independent of $Q$ 's sign and size ✓ — the trap in the question.

Recall Quick self-test

Two equal $+ q$ at $\pm a$ , field on the axis of symmetry points which way? ::: Along the axis (away from the charges), horizontal parts cancel. Dipole on the perpendicular bisector points which way? ::: Antiparallel to the dipole (from $+$ to $-$ side horizontally), vertical parts cancel. Between two unequal like charges, is the null nearer the bigger or smaller? ::: Nearer the smaller; distance ratio is $q_{1} / q_{2}$ . Why is a ring's centre field zero? ::: Opposite pieces cancel by symmetry, for any uniform $Q$ and $R$ . A negative charge hovers; which way does $E$ point? ::: Downward — force $q E$ is opposite $E$ , so upward force needs downward field.

Parent: topic note.