Electric field — definition, field lines, superposition: Exercises

Level 1 — Recognition

Can you pick the right formula and read off signs/directions?

Recall Solution L1·1

WHAT we use: the field of a point charge, $E = \frac{k ∣ q ∣}{r ^{2}}$ . WHY this formula: a single isolated point charge is exactly the case Coulomb's law was built for; dividing the Coulomb force by the test charge left $E = k ∣ q ∣/ r^{2}$ .

Plug in ( $3.0 nC = 3.0 \times 1 0^{- 9} C$ , $2.0 cm = 0.020 m$ ): $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 9} )}{( 0.020 ) ^{2}} = \frac{26.97}{4.0 \times 1 0 ^{- 4}} \approx 6.74 \times 1 0^{4} N/C .$ Direction: $q$ is positive, so $E$ points radially outward, away from the origin.

Recall Solution L1·2

WHAT: $E = \frac{k q}{r ^{2}} \overset{r}{^}$ , where $\overset{r}{^}$ points from the source to the field point — here that is the $+ x$ direction. WHY the sign matters: with $q$ negative, $\frac{k q}{r ^{2}}$ is a negative number times $\overset{r}{^}$ , so $E$ points opposite to $\overset{r}{^}$ — i.e. back toward the charge, in the $- x$ direction. What it looks like: field lines enter a negative charge, so at any nearby point the arrow points inward. ✓

Level 2 — Application

Plug numbers into a multi-step but still single-idea calculation.

Recall Solution L2·1

WHAT: once you know $E$ , the force on any charge is $F = Q E$ . WHY this is one line: the field already packaged everything about the sources; the charge just responds. $∣ F ∣ = ∣ - e ∣ E = (1.6 \times 1 0^{- 19}) (2.0 \times 1 0^{4}) = 3.2 \times 1 0^{- 15} N .$ Direction: $Q = - e$ is negative, so $F$ points opposite to $E$ → in the $- x$ direction.

Recall Solution L2·2

WHAT we need: a point where the two fields have equal magnitude and opposite direction so they cancel. WHY between them: between two positive charges the field from the left one points right, from the right one points left — only here can they oppose and cancel.

Let the null point be at distance $x$ from $+ q$ , so it is $d - x$ from $+ 4 q$ . Set magnitudes equal: $\frac{k q}{x ^{2}} = \frac{k ( 4 q )}{( d - x ) ^{2}} \Rightarrow \frac{1}{x ^{2}} = \frac{4}{( d - x ) ^{2}} .$ Take the square root (both sides positive): $\frac{1}{x} = \frac{2}{d - x} \Rightarrow d - x = 2 x \Rightarrow x = \frac{d}{3} .$ What it looks like: the null sits closer to the smaller charge — makes sense, you must stand near the weaker charge for its field to keep up with the stronger one.

Level 3 — Analysis

Now the geometry does the work — components and symmetry.

Figure — Electric field — definition, field lines, superposition

Recall Solution L3·1

Step 1 — distance (look at the triangle in the figure). Each charge is a distance $r = a^{2} + y^{2}$ from the field point (Pythagoras on the right triangle with legs $a$ and $y$ ). So each field has magnitude $E_{1} = \frac{k q}{a ^{2} + y ^{2}}$ .

Step 2 — components (WHY symmetry helps). The two charges mirror each other across the $y$ -axis. Their horizontal parts $E_{x}$ point in opposite directions and cancel; their vertical parts both point $+ y$ and add. The vertical fraction is $cos θ = \frac{y}{a ^{2} + y ^{2}}$ — the "adjacent over hypotenuse" of that same triangle.

Step 3 — total. $E_{y} = 2 E_{1} cos θ = \frac{2 k q}{a ^{2} + y ^{2}} \cdot \frac{y}{a ^{2} + y ^{2}} = \frac{2 k q y}{( a ^{2} + y ^{2} ) ^{3/2}} .$ The $3/2$ power: one full power of $r^{2}$ from Coulomb, plus half a power from the projection $cos θ$ .

Numbers ( $q = 2.0 \times 1 0^{- 9}$ , $a = 0.03$ , $y = 0.04$ , so $a^{2} + y^{2} = 0.0025 m^{2}$ , $(a^{2} + y^{2})^{3/2} = 1.25 \times 1 0^{- 4}$ ): $E_{y} = \frac{2 ( 8.99 \times 1 0 ^{9} ) ( 2.0 \times 1 0 ^{- 9} ) ( 0.04 )}{1.25 \times 1 0 ^{- 4}} \approx 1.15 \times 1 0^{4} N/C, pointing + y .$

Recall Solution L3·2

What changed: with one charge negative, the vertical parts now cancel and the horizontal parts add (both point in $- x$ ). We project with $cos ϕ = \frac{a}{a ^{2} + y ^{2}}$ (the other leg of the triangle). $E = 2 E_{1} cos ϕ = \frac{2 k q}{a ^{2} + y ^{2}} \cdot \frac{a}{a ^{2} + y ^{2}} = \frac{2 k q a}{( a ^{2} + y ^{2} ) ^{3/2}}, along - x .$ This is the Electric Dipole result; defining $p = 2 q a$ , far away it becomes $E \to \frac{k p}{y ^{3}}$ .

Level 4 — Synthesis

Combine superposition, limits, and physical reasoning.

Recall Solution L4·1

WHAT the limit does: when $y ≫ a$ , the $a^{2}$ inside $(a^{2} + y^{2})^{3/2}$ is negligible next to $y^{2}$ , so $(a^{2} + y^{2})^{3/2} \to (y^{2})^{3/2} = y^{3}$ . $E_{y} \to \frac{2 k q y}{y ^{3}} = \frac{2 k q}{y ^{2}} = \frac{k ( 2 q )}{y ^{2}} .$ WHY this is exactly right: from far away the two charges blur into one blob of total charge $2 q$ — you can no longer resolve the $\pm a$ separation. The inverse-square law of a single charge re-emerges. Number at $y = 1.0 m$ : $E \approx \frac{( 8.99 \times 1 0 ^{9} ) ( 4.0 \times 1 0 ^{- 9} )}{1. 0 ^{2}} \approx 36 N/C$ .

Recall Solution L4·2

$E \to \frac{2 k q a}{y ^{3}} = \frac{k p}{y ^{3}}, p = 2 q a .$ WHY steeper: the total charge of a dipole is zero. From far away the $+ q$ and $- q$ nearly cancel, so the leading $1/ y^{2}$ term vanishes. What survives is only the tiny effect of their separation $2 a$ — a second-order leftover that decays one power faster, $1/ y^{3}$ . What it looks like: the field lines of a dipole loop tightly from $+$ to $-$ ; far out there are almost none, because the outgoing lines from $+ q$ are captured by $- q$ nearby.

Level 5 — Mastery

One problem that forces every tool together.

Recall Solution L5·1

Step 1 — chop the ring (Continuous charge, look at the figure). Split the ring into tiny pieces $d q$ . Each piece is a distance $r = a^{2} + z^{2}$ from the axial point (Pythagoras: leg $a$ from centre to the piece, leg $z$ along the axis). Each produces a field $d E = \frac{k d q}{a ^{2} + z ^{2}}$ . This is Continuous Charge Distributions + Superposition Principle working together.

Step 2 — symmetry kills the sideways parts. For every piece there is an opposite piece across the ring. Their components perpendicular to the axis cancel; only the axial component survives. The axial fraction is $cos θ = \frac{z}{a ^{2} + z ^{2}}$ . $d E_{z} = \frac{k d q}{a ^{2} + z ^{2}} \cdot \frac{z}{a ^{2} + z ^{2}} = \frac{k z d q}{( a ^{2} + z ^{2} ) ^{3/2}} .$

Step 3 — add up (integrate). Everything multiplying $d q$ is the same for every piece (all pieces share the same $r$ and $z$ ), so $\int d q = Q$ : $E_{z} = \frac{k Q z}{( a ^{2} + z ^{2} ) ^{3/2}} directed along the axis.$

Numbers ( $Q = 8.0 \times 1 0^{- 9}$ , $a = 0.05$ , $z = 0.12$ ; $a^{2} + z^{2} = 0.0025 + 0.0144 = 0.0169$ , $(0.0169)^{3/2} = 2.197 \times 1 0^{- 3}$ ): $E_{z} = \frac{( 8.99 \times 1 0 ^{9} ) ( 8.0 \times 1 0 ^{- 9} ) ( 0.12 )}{2.197 \times 1 0 ^{- 3}} \approx 3.93 \times 1 0^{3} N/C .$

Far-field check ( $z ≫ a$ ): $(a^{2} + z^{2})^{3/2} \to z^{3}$ , so $E_{z} \to \frac{k Q z}{z ^{3}} = \frac{k Q}{z ^{2}}$ — a point charge $Q$ , as it must be. ✓ Centre check ( $z = 0$ ): $E_{z} = 0$ — at the exact centre every piece's field is cancelled by its opposite. ✓

Recall One-line summary of the ladder

Recognise the point-charge formula → apply $F = Q E$ and find nulls → analyse with components and symmetry → synthesise via far-field limits → master the continuous ring by chop-project-add.