Visual walkthrough — Electric field — definition, field lines, superposition

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Step 1 — Draw the stage: two charges, one point to measure

WHAT. Put a positive charge $+q$ on the flat ground line at position $+a$ to the right of centre, and an identical $+q$ at $-a$ to the left. We want to know the electric field at a point $P$ that floats straight above the centre, at height $y$.

WHY these choices. We pick a point on the vertical centre-line because the setup is a mirror image left-to-right. Symmetry will do half our work for free later — but only if we sit exactly on the mirror line. If we sat off to one side, nothing would cancel and the algebra would explode.

PICTURE. Look at the figure. The two red dots are the charges. The green dot is $P$. The letter $a$ is the horizontal distance from centre to each charge; $y$ is the straight-up height of $P$.

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Step 2 — How far is each charge from $P$? (Pythagoras)

WHAT. Find the straight-line distance $r$ from one charge (say the right one) to $P$. That charge sits at ground-level a distance $a$ sideways, and $P$ is $y$ up. Those two legs meet at a right angle, so the diagonal is $r=\underset{\text{diagonal of the right triangle}}{\underbrace{\sqrt{a^2+y^2}}}.$

WHY the square root of a sum of squares. The horizontal leg ($a$) and the vertical leg ($y$) form the two short sides of a right triangle; $r$ is its hypotenuse. Pythagoras (${\text{leg}}^2+{\text{leg}}^2={\text{hyp}}^2$) is the only tool that turns two perpendicular distances into one straight distance — that is exactly the question we are asking.

PICTURE. The blue triangle: bottom leg $a$, vertical leg $y$, slanted hypotenuse $r$ running from the right charge up to $P$. By left-right symmetry the left charge has an identical triangle, mirror-flipped — so both charges are the same distance $r$ from $P$.

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Step 3 — Strength of ONE charge's field (Coulomb's Law)

WHAT. From Coulomb's Law the field made by a single point charge, at distance $r$, has magnitude $E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}=k\frac{q}{r^2}=\frac{kq}{a^2+y^2}.$ Read it piece by piece:

• $k=\frac{1}{4\pi {\epsilon }_0}\approx 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ ::: the universal strength constant.
• $q$ ::: bigger charge → stronger field (top of the fraction).
• $r^2=a^2+y^2$ ::: farther away → weaker field, and it weakens as the square of distance (bottom).

WHY squared, not just $r$? A point charge sends its field out equally in all directions, spreading over a sphere. A sphere's surface area is $4\pi r^2$. The same "amount of field" thinned over a larger sphere means strength $\propto 1/r^2$. That is why the exponent is 2, and it comes straight from geometry.

PICTURE. From each red charge an arrow $E_1$ points directly away (both charges are positive, so they push a positive test charge outward). Each arrow lies along its own hypotenuse line. Both arrows have the same length because both $r$'s are equal.

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Step 4 — A slanted arrow is really two arrows (split into $x$ and $y$)

WHAT. Each $E_1$ arrow points diagonally. We break it into a horizontal piece $E_{1x}$ and a vertical piece $E_{1y}$. Using the same blue triangle, the angle $\theta$ is measured at $P$ between the vertical and the hypotenuse, and $E_{1y}=E_1\cos \theta ,E_{1x}=E_1\sin \theta .$

WHY cosine here? Cosine answers one specific question: "of this slanted length, how much lies along the direction I care about (the vertical)?" It is defined as $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$. The side adjacent to the angle $\theta$ (the vertical leg) has length $y$, and the hypotenuse is $r$, so $\cos \theta =\frac{y}{r}=\frac{y}{\sqrt{a^2+y^2}}.$ We use cosine and not sine because the piece we ultimately want (the vertical) sits next to the angle, not across from it.

PICTURE. The diagonal arrow is redrawn as an L-shape: a green vertical component going up, a yellow horizontal component going sideways. The angle $\theta$ is marked at $P$ between the vertical dashed line and the arrow.

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Step 5 — Add the two fields (Superposition) and watch the horizontals die

WHAT. By the Superposition Principle the total field is the plain vector sum of the two. Add the pieces separately: $E_x^{\text{tot}}=E_{1x}^{(\text{left})}+E_{1x}^{(\text{right})},E_y^{\text{tot}}=E_{1y}^{(\text{left})}+E_{1y}^{(\text{right})}.$

• Horizontals cancel. The right charge pushes $P$ up-and-to-the-left? No — up-and-to-the-right (away from itself). The left charge pushes up-and-to-the-left. Those two horizontal pieces are equal in size and opposite in direction, so $E_x^{\text{tot}}=0$.
• Verticals add. Both charges push $P$ upward (both are below $P$ and repel it upward). Same size, same direction → they double.

$E_y^{\text{tot}}=2E_{1y}=2E_1\cos \theta .$

WHY the cancellation is guaranteed (not luck). The whole picture is a mirror reflection about the vertical line through $P$. Reflection flips left↔right, so it flips the sign of every horizontal component but leaves verticals untouched. A quantity that must equal its own negative can only be zero. That is the reason $E_x=0$ — pure symmetry, no arithmetic needed.

PICTURE. The two diagonal arrows placed tail-to-tail at $P$; their yellow horizontals point opposite ways and erase each other (shown greyed), while the green verticals stack into one tall arrow.

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Step 6 — Assemble the final formula

WHAT. Substitute $E_1$ and $\cos \theta$ into $E_y^{\text{tot}}=2E_1\cos \theta$:

=\frac{2kqy}{(a^2+y^2)^{3/2}}.$$ Track where the exponent $3/2$ is born: - one full power $(a^2+y^2)^{1}$ from Coulomb's $r^2$ in $E_1$, - half a power $(a^2+y^2)^{1/2}$ from the $\sqrt{\,}$ in $\cos\theta$, - together $(a^2+y^2)^{1+1/2}=(a^2+y^2)^{3/2}$. **WHY that mixed power matters.** It is *not* a pure inverse-square. The extra half-power comes from the geometry of projecting the slant onto the vertical — a signature you will meet again in every "on-axis of a symmetric arrangement" problem. **PICTURE.** The clean result: a single upward green arrow at $P$, its length labelled by the boxed formula, the two charges below. ![[deepdives/dd-physics-1.8.04-d2-s06.png]] > [!formula] The central result of this walkthrough > $$\boxed{\,E_y=\frac{2kqy}{(a^2+y^2)^{3/2}}\,}\qquad(\text{pointing straight up, }E_x=0)$$ --- ## Step 7 — The edge and degenerate cases (never leave the reader stranded) **Case A — very far away, $y\gg a$.** When $y$ dwarfs $a$, the term $a^2$ under the root is a crumb next to $y^2$; drop it: $a^2+y^2\approx y^2$. Then $$E_y\approx\frac{2kqy}{(y^2)^{3/2}}=\frac{2kqy}{y^3}=\frac{2kq}{y^2}=\frac{k(2q)}{y^2}.$$ From far off the two charges blur into **one lump of charge $2q$** — an ordinary inverse-square point-charge field. *Forecast confirmed:* far away, only the **total** charge matters, not its spread. This is why very distant charge clusters look like [[Continuous Charge Distributions]] collapsed to a point. **Case B — right at the centre, $y=0$.** Plug $y=0$: $$E_y=\frac{2kq\cdot 0}{(a^2)^{3/2}}=0.$$ At the exact midpoint between two equal positive charges, the left charge pushes right and the right charge pushes left with equal strength — total field **zero**. This is a *balance point* (an unstable one: nudge a test charge sideways and it runs off). See the [[Electric Field|field-line]] saddle. **Case C — the opposite arrangement (a hint of the [[Electric Dipole]]).** If the right charge were $-q$ instead of $+q$, the **verticals** would cancel and the **horizontals** would survive — the mirror image of everything above. That flipped problem gives the dipole's $1/y^3$ field. Same triangle, roles of $\sin$ and $\cos$ swapped. **PICTURE.** Three mini-panels: (A) far field arrow shrinking as one point charge, (B) zero arrow at the origin with two opposing pushes, (C) the flipped dipole case with a sideways arrow. ![[deepdives/dd-physics-1.8.04-d2-s07.png]] > [!mistake] "Just add the magnitudes: $E=E_1+E_1=2E_1$." > That naked "2" ignores that the arrows are **slanted**. The real total is $2E_1\cos\theta$, always *smaller* than $2E_1$ because $\cos\theta\le 1$. The factor $\cos\theta$ is the price of the horizontal pieces throwing themselves away. --- ## The one-picture summary Everything on one canvas: the two triangles, the two slanted $E_1$ arrows, the horizontals annihilating, the verticals stacking, and the final boxed formula with each factor colour-tagged to where it was born. ![[deepdives/dd-physics-1.8.04-d2-s08.png]] > [!recall]- Feynman retelling — the whole walk in plain words > Two identical positive balls sit on the ground, one to the left, one to the right of a centre spot. We hover a "field-feeler" straight above the centre. **Step 1–2:** each ball is the same slanted distance away — draw the right triangle, its hypotenuse is $\sqrt{a^2+y^2}$. **Step 3:** each ball's push weakens with the *square* of that distance (spreading over a sphere). **Step 4:** each push points diagonally away from its ball, so we split it into a sideways part and an up part; the up part is the whole times $\cos\theta=y/r$. **Step 5:** the two sideways parts point opposite ways and cancel (the picture is a mirror), while the two up-parts point the same way and double. **Step 6:** double the up-part gives $\dfrac{2kqy}{(a^2+y^2)^{3/2}}$ — the odd $3/2$ power is one power from Coulomb plus half a power from the projection. **Step 7:** stand far away and the pair looks like one charge $2q$; stand at the exact middle and the pushes tie, so the field is zero. Flip one charge's sign and the whole story turns sideways — that is the dipole. > [!recall]- Quick self-test > Why is $E_x=0$ on the centre-line? ::: Mirror symmetry flips horizontal components; a quantity equal to its own negative must be zero. > Where does the $3/2$ exponent come from? ::: $1$ power of $(a^2+y^2)$ from Coulomb's $r^2$ plus $1/2$ power from $\cos\theta$'s square root. > What does $E_y$ become for $y\gg a$? ::: $k(2q)/y^2$ — a single point charge of the combined charge. > What is $E_y$ exactly at $y=0$? ::: Zero — the two equal pushes cancel at the midpoint.