Visual walkthrough — Potential of point charge, potential from field and vice versa
Step 1 — The players: a charge, a test charge, a distance
WHAT. We have one fixed charge sitting at a point. Around it lives an invisible influence. To feel that influence we carry a second, very tiny positive charge — call it — to some spot a distance away.
WHY. "Potential" is defined by what an outside hand must do to move around. So we must have in the picture and a way to measure how far out we are: that distance is (in metres).
WHAT IT LOOKS LIKE. Look at the figure. The purple dot is the source charge . The coral dot is our test charge . The dashed line between them is — the only geometric quantity the answer will depend on, because everything here is round (symmetric) about .
Step 2 — The force per charge: the field
WHAT. At distance , the source charge pushes our test charge with a force. Divide that force by and you get the force per unit charge — the electric field . From Coulomb's Law this is:
Reading it left to right: is the arrow we want; is just a units constant; says the push weakens as the square of distance; (a length-one arrow) says the push points radially outward from a positive .
WHY. Potential is defined through the field, so we must know the field first. This is our one starting fact — see Electric Field of Point Charge.
WHAT IT LOOKS LIKE. The figure shows arrows fanning outward. Notice they get shorter as grows — that shortening is the law drawn.
Step 3 — Potential is work per charge, so we must add up a push along a path
WHAT. To bring from very far away (infinity) inward to distance , an outside hand fights the field the whole trip. The total work the field does, divided by , defines potential:
Term by term: is the number we hunt (in volts); the big means "add up over the whole trip"; is one tiny step of the journey; is how much the field helps or fights during that step; the minus sign flips field's work into hand's work.
WHY THIS TOOL — the integral? A single push is only true at one distance; the trip crosses infinitely many distances, each with a different push. An integral is exactly the tool that sums infinitely many tiny contributions that keep changing. A plain multiplication would only work if the push were constant — it is not.
WHY the minus sign? The hand pushes against the field. Work by the hand (work by the field). Dividing by keeps that minus.
WHAT IT LOOKS LIKE. The figure chops the incoming path into little segments ; each segment is a small slice of work. We stack the slices to get .
Step 4 — Choose the smartest path: straight in along
WHAT. We are free to travel any route from to , because the electrostatic field is conservative (path doesn't matter — see Conservative Fields and Curl). So pick the easiest route: march straight inward along the radial line. Then the step is and the dot product collapses:
Here is a moving distance (a dummy variable) that slides from down to the final ; we rename it so it doesn't clash with the endpoint .
WHY. On a radial path, field and motion are perfectly parallel, so the messy angle disappears and becomes an ordinary 1-D integral we can actually compute.
WHAT IT LOOKS LIKE. In the figure the path (coral) lies exactly on top of a field arrow (lavender) — they overlap. Compare the greyed-out curvy path: it gives the same answer but is harder to compute. That equality is the whole gift of a conservative field.
Step 5 — Do the integral (the calculus punchline)
WHAT. Substitute the field and integrate:
\;=\; -\frac{q}{4\pi\varepsilon_0}\int_{\infty}^{r} r'^{-2}\,dr'$$ We pull the constants out front. The remaining piece uses the single fact that the antiderivative of $r'^{-2}$ is $-\,r'^{-1}$: $$\int r'^{-2}\,dr' = -\frac{1}{r'}\quad\Longrightarrow\quad V(r) = -\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r}$$ **WHY THIS TOOL — antiderivative of $1/r'^2$?** The integral asks "what function, when differentiated, gives $1/r'^2$?" That is $-1/r'$. This is the *reverse* of the very derivative we'll take in Step 7 — the bridge is already visible. **WHAT IT LOOKS LIKE.** The figure plots $1/r'^2$; the shaded area under it from $\infty$ down to $r$ **is** the integral. The area is finite even though the tail runs to infinity — because $1/r'^2$ shrinks fast enough. That's why potential is a finite number. --- ## Step 6 — Plug in the limits: the reference at infinity vanishes **WHAT.** Evaluate the bracket at the two ends: $$V(r) = -\frac{q}{4\pi\varepsilon_0}\left(-\frac{1}{r} - \Big(-\underbrace{\frac{1}{\infty}}_{\to\,0}\Big)\right) = -\frac{q}{4\pi\varepsilon_0}\left(-\frac{1}{r}\right) = \boxed{\;\frac{1}{4\pi\varepsilon_0}\frac{q}{r} = \frac{kq}{r}\;}$$ At $r'=\infty$ the term $1/r'\to 0$ — that is the whole reason we chose infinity as our starting point: it makes the far end contribute **nothing**, so $V=0$ infinitely far away. Two minus signs then cancel, leaving a clean positive answer for positive $q$. **WHY.** A potential is only defined *relative* to a reference. Choosing $V(\infty)=0$ is the choice that kills one term and gives the tidy formula. **WHAT IT LOOKS LIKE.** The figure draws $V(r)=kq/r$: a curve that dives steeply near the charge and flattens toward zero far away. Overlaid faintly is $E\propto 1/r^2$ — note $V$ falls off **slower** (one power of $r$, not two). > [!formula] The result > $$V(r)=\frac{kq}{r},\qquad k=\frac{1}{4\pi\varepsilon_0}\approx 9\times10^{9}\ \mathrm{N\,m^2/C^2}$$ --- ## Step 7 — Cross-check by walking back: $\vec E$ from $V$ **WHAT.** The bridge should work both directions. From $dV=-\vec E\cdot d\vec l$, a small radial step gives $E_r=-\dfrac{dV}{dr}$. Feed in our fresh result: $$E_r = -\frac{d}{dr}\!\left(\frac{kq}{r}\right) = -kq\cdot\left(-\frac{1}{r^2}\right) = \frac{kq}{r^2}\quad\checkmark$$ We got the **original** field back. The derivative in Step 7 exactly undoes the integral of Step 5. **WHY THIS TOOL — the derivative $-dV/dr$?** Potential is a "height landscape." The field is its steepest slope. A derivative *is* slope, and the minus sign makes it point **downhill** — toward lower potential. In full 3-D this is $\vec E=-\nabla V$ (see [[Gradient Operator]]). **WHAT IT LOOKS LIKE.** The figure shows the $V$ curve as a hill; at one point a tangent line is drawn, and a red arrow marks the field pointing **down** that slope. Steep part of the hill ⇒ long arrow ⇒ strong field. --- ## Step 8 — Sign & degenerate cases (never leave a gap) **WHAT.** Walk through what the formula does at every extreme: - **$q>0$ (positive charge):** $V=kq/r>0$ everywhere. Landscape is a **hill** rising to $+\infty$ at the charge. - **$q<0$ (negative charge):** $V=kq/r<0$ everywhere. Landscape is a **funnel/well** plunging to $-\infty$. - **$r\to 0$ (right at the charge):** $V\to\pm\infty$. This is an *idealisation* — real charges have finite size, so the blow-up is not physical. - **$r\to\infty$:** $V\to 0$, matching our reference choice. - **Field zero but $V\neq 0$:** at the midpoint between two equal positive charges the two $\vec E$ arrows cancel, yet both potentials **add** (scalars) to a large positive $V$. Potential depends on the *whole trip*, not the local push. **WHY.** A reader must never meet a sign or a limiting case you skipped. **WHAT IT LOOKS LIKE.** The figure stacks the positive-charge hill (above zero) and the negative-charge well (below zero) on one axis, with the flat $V=0$ line far out marked in the middle. > [!mistake] "$V=0$ wherever $\vec E=0$." > The two are linked by an integral, not by equality. Cancelling arrows ($\vec E=0$) can still sit on top of a big height ($V\neq0$). See [[Equipotential Surfaces]] for the flip side — flat height means no field. --- ## The one-picture summary The whole journey on one canvas: a source charge, a radial path inward from infinity, the shrinking $1/r^2$ field arrows we summed, the finite area they enclose, and the resulting $V=kq/r$ hill — with the field as its downhill slope. Read left→right for the derivation; right→left for the check. > [!recall]- Feynman retelling — say it to a friend > Picture one glowing dot (the charge) and you holding a tiny bead (the test charge) far, far away. As you walk the bead inward, the dot pushes back — gently far off, harder up close (that's the $1/r^2$ arrows getting shorter with distance). "Potential" is just the *total effort per unit bead* it took to make that walk — you add up every little push along the way, which is what the integral does. Because it doesn't matter which path you took, we walk straight in to keep the sum easy. Summing $1/r'^2$ from infinity down to $r$ gives a neat $1/r$ — that's $V=kq/r$, a hill that's tall near the dot and flat far away. And if someone hands you the hill instead, you find the field by reading its slope and pointing downhill — that's $\vec E=-dV/dr$. One derivation, walkable both ways. > [!recall]- Two-line self-test > Why does the $\infty$ end of the integral drop out? ::: Because $1/r'\to 0$ there — that's the whole point of choosing $V(\infty)=0$. > Why does integrating $1/r^2$ give a $1/r$ potential? ::: The antiderivative of $r^{-2}$ is $-r^{-1}$, so one power of $r$ survives in the denominator. --- ## Connections - [[Coulomb's Law]] — supplies the $1/r^2$ field we integrated. - [[Electric Field of Point Charge]] — recovered exactly in Step 7. - [[Gradient Operator]] — the 3-D machinery behind $\vec E=-\nabla V$. - [[Conservative Fields and Curl]] — why we could pick the radial shortcut in Step 4. - [[Equipotential Surfaces]] — the flat-height view of "no field." - [[Potential Energy of Charge System]] — multiply $V$ by a charge to get energy. --- #flashcards/physics Why can we choose a radial path when computing $V$ of a point charge? ::: The field is conservative, so the line integral is path-independent — pick the easiest (radial) route. What makes the infinity limit vanish in the integral? ::: $1/r'\to 0$ as $r'\to\infty$, chosen so that $V(\infty)=0$. Why is $V$ finite even though the path starts at infinity? ::: $1/r'^2$ shrinks fast enough that the area under it is finite.