Potential of point charge, potential from field and vice versa

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1. What is potential? (definition first)

Why the minus sign? The external agent must push against the field's force $q_0\vec E$. Work by the agent $= -\,$work by the field $= -\int q_0\vec E\cdot d\vec l$. Divide by $q_0$.

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2. Deriving $V$ from $\vec{E}$ — the point charge (from scratch)

We compute $V$ at distance $r$ from charge $q$, bringing the test charge in radially from $\infty$.

Step 1 — set up the line integral. $V(r) = -\int_{\infty}^{r}\vec{E}\cdot d\vec{l}$ Why this step? This is the definition of potential; we just plug our known $\vec E$ in.

Step 2 — choose a radial path. Move straight inward, so $d\vec{l} = dr'\,\hat r$ and $\vec E\cdot d\vec l = E\,dr'$. Why this step? Electrostatic $V$ is path-independent (field is conservative), so we pick the easiest path — pure radial — and the angular part vanishes.

Step 3 — substitute and integrate.

= -\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r}$$ *Why this step?* $\int r'^{-2}dr' = -r'^{-1}$. Evaluate the bracket. **Step 4 — evaluate limits.** At $\infty$, $1/r'\to 0$; at $r$ we get $1/r$. $$\boxed{V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}}$$ > [!formula] Potential of a point charge > $$V(r) = \frac{kq}{r},\qquad k=\frac{1}{4\pi\varepsilon_0}\approx 9\times 10^{9}\ \mathrm{N\,m^2/C^2}$$ > Note: $V\propto 1/r$ (potential), while $E\propto 1/r^2$ (field). The potential falls off **slower**. ![[1.8.09-Potential-of-point-charge,-potential-from-field-and-vice-versa.png]] --- ## 3. Going back: $\vec{E}$ from $V$ (the gradient) The integral inverts to a derivative. Starting from $dV = -\vec E\cdot d\vec l$, for a small radial step $dV = -E_r\,dr$, so $$E_r = -\frac{dV}{dr}$$ In full 3-D the field is the **negative gradient**: $$\boxed{\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat x + \frac{\partial V}{\partial y}\hat y + \frac{\partial V}{\partial z}\hat z\right)}$$ **Check it on the point charge:** with $V=kq/r$, $$E_r = -\frac{d}{dr}\!\left(\frac{kq}{r}\right) = -kq\cdot\left(-\frac{1}{r^2}\right) = \frac{kq}{r^2}\ \checkmark$$ We recovered the original field — the bridge works both ways. > [!intuition] Slope picture > $V$ is a "landscape." $\vec E$ points in the direction of **steepest descent** and its magnitude is the **steepness**. Flat region (constant $V$) ⇒ no field. Where contour lines crowd together ⇒ strong field. --- ## 4. Why $V$ is path-independent (the deep reason) For ANY closed loop, $\oint \vec E\cdot d\vec l = 0$ in electrostatics (curl-free field). So the work to go $A\to B$ is the same on every path; only endpoints matter. This is exactly what lets us *define* a single-valued $V(P)$. If it weren't true, "potential" wouldn't be a well-defined function. --- ## 5. Worked examples > [!example] (a) Potential of two charges (scalar superposition) > Charges $q_1=+2\ \mu C$ at origin, $q_2=-3\ \mu C$ at $x=4\ \mathrm m$. Find $V$ at $x=1\ \mathrm m$. > > **Step 1.** Distances: from $q_1$, $r_1=1\,\mathrm m$; from $q_2$, $r_2=3\,\mathrm m$. > *Why?* $V$ depends only on distance, so just measure each. > > **Step 2.** Add scalars (signs included): > $$V = k\frac{q_1}{r_1}+k\frac{q_2}{r_2}=9\times10^9\!\left(\frac{2\times10^{-6}}{1}+\frac{-3\times10^{-6}}{3}\right)$$ > *Why?* Potential superposes algebraically — no components, sign of charge carried directly. > > **Step 3.** Compute: $9\times10^9(2\times10^{-6}-1\times10^{-6})=9\times10^9\times10^{-6}=9000\ \mathrm V$. > [!example] (b) Field from a given potential > $V(x)=3x^2-2x$ (volts, $x$ in metres). Find $E_x$ at $x=2$. > > **Step 1.** $E_x=-\dfrac{dV}{dx}=-(6x-2)$. > *Why?* Field is minus the slope of the potential. > > **Step 2.** At $x=2$: $E_x=-(12-2)=-10\ \mathrm{V/m}$ (points in $-x$). > *Why the sign?* $V$ is rising with $x$ here, so the field points the other way — downhill. > [!example] (c) Potential difference from field > Uniform field $E=200\ \mathrm{V/m}$ along $+x$. Find $V_A-V_B$ between $x_A=1$ and $x_B=4\,\mathrm m$. > > **Step 1.** $V_A-V_B=-\int_B^A E\,dx = -E\,(x_A-x_B)=-200(1-4)=600\ \mathrm V$. > *Why?* Moving against the field (from high $x$ to low $x$) raises potential; $A$ is at higher $V$. --- ## 6. Common mistakes (steel-manned) > [!mistake] "$V$ is zero where $\vec E$ is zero." > **Why it feels right:** in many simple cases (e.g., far away) both vanish together. > **The fix:** $V$ depends on the *integral* of $E$, not its local value. At the **midpoint of two equal positive charges**, $\vec E=0$ by symmetry but $V$ is large and positive. Conversely, between $+q$ and $-q$ midpoint, $V=0$ but $\vec E\neq 0$. > [!mistake] "Add fields as scalars like potentials." > **Why it feels right:** superposition works for both. > **The fix:** $\vec E$ is a **vector** — add components. $V$ is a **scalar** — add (signed) magnitudes. Mixing them gives wrong angles and wrong magnitudes. > [!mistake] "Drop the minus sign in $\vec E=-\nabla V$." > **Why it feels right:** sign feels like bookkeeping. > **The fix:** without it, field would point *uphill* and energy conservation breaks. The minus encodes "force points toward lower potential energy." --- ## 7. Active recall > [!recall]- Quick self-test (cover answers) > - Why is $V$ a scalar but $\vec E$ a vector? → $V$ is energy/charge (no direction); $\vec E$ is force/charge. > - $V\propto ?$ and $E\propto ?$ for a point charge? → $V\propto 1/r$, $E\propto 1/r^2$. > - How to get $\vec E$ from $V$? → $\vec E=-\nabla V$. > - Can $V\neq0$ where $\vec E=0$? → Yes (midpoint of two like charges). > [!recall]- Feynman: explain to a 12-year-old > Imagine a hilly field. The **height** of the ground at each spot is the *potential* $V$ — just a number telling how "high up" you are. A ball placed there feels a push — that push is the *electric field* $\vec E$, and it always points **down the steepest slope**. If the ground is flat, the ball feels no push (no field). Steep cliff ⇒ strong push. So if I give you the shape of the hills, you know which way every ball rolls and how hard — that's getting field from potential. And if I tell you how hard balls are pushed everywhere, you can add up the climbs to rebuild the hill heights — that's getting potential from field. > [!mnemonic] > **"Volts go down, Fields go down the slope."** > $V$ uses $1/r$ (one $r$ on the bottom = "one"); $E$ uses $1/r^2$ ("squared, stronger drop"). And **E = minus slope of V**: *Eat Minus the Slope*. --- ## 8. Connections - [[Coulomb's Law]] — the $1/r^2$ field we integrated. - [[Electric Field of Point Charge]] — what we recover from $-\nabla V$. - [[Potential Energy of Charge System]] — $U=qV$. - [[Equipotential Surfaces]] — surfaces of constant $V$, always $\perp \vec E$. - [[Conservative Fields and Curl]] — why $\oint\vec E\cdot d\vec l=0$. - [[Gradient Operator]] — the math machinery of $-\nabla V$. --- #flashcards/physics Define electric potential at a point ::: Work done per unit positive test charge by an external agent to bring it from infinity to that point: $V=-\int_\infty^P \vec E\cdot d\vec l$. Unit: volt (J/C). Potential of a point charge formula ::: $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}=\dfrac{kq}{r}$. How does $V$ relate to $\vec E$ (integral form)? ::: $V=-\int \vec E\cdot d\vec l$ (negative line integral of field). How to get $\vec E$ from $V$? ::: $\vec E=-\nabla V$; in 1-D $E_x=-dV/dx$. Why the minus sign in $\vec E=-\nabla V$? ::: Field points toward decreasing potential (downhill), encoding that force lowers potential energy. $V$ vs $r$ and $E$ vs $r$ scaling for a point charge ::: $V\propto 1/r$, $E\propto 1/r^2$; potential falls off slower. Is $V$ a scalar or vector, and how do you superpose it? ::: Scalar; add algebraically with signs: $V=\sum kq_i/r_i$. Can the field be zero where potential is nonzero? ::: Yes — e.g. midpoint of two equal positive charges: $\vec E=0$ but $V>0$. Why is electrostatic potential path-independent? ::: Because $\oint\vec E\cdot d\vec l=0$ (field is conservative/curl-free); work depends only on endpoints. Recover $E$ from $V=kq/r$ ::: $E_r=-dV/dr=-kq(-1/r^2)=kq/r^2$. ✓ ## 🖼️ Concept Map ```mermaid flowchart TD E["Field E vector"] V["Potential V scalar"] DEF["V = work per unit charge"] LINT["V = -∫ E·dl"] GRAD["E = -∇V"] PC["Point charge field E = kq/r²"] VPC["V = kq/r"] RAD["Radial path choice"] CONS["Conservative field"] SLOPE["Field points downhill"] DEF -->|leads to| LINT E -->|line integral| LINT LINT -->|defines| V V -->|negative gradient| GRAD GRAD -->|recovers| E PC -->|integrate radially| VPC CONS -->|allows| RAD RAD -->|simplifies| LINT LINT -->|applied to point charge| VPC VPC -->|falls off as 1 over r| V GRAD -->|explains| SLOPE ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, do cheezein hain jo aksar confuse karti hain: **field** $\vec E$ aur **potential** $V$. Field batata hai ki kisi point par charge par kitna aur kis direction me force lagega (force per charge, vector). Potential batata hai ki us point tak ek test charge ko laane me kitni energy lagti hai (energy per charge, scalar). Scalar hone ki wajah se potential ka kaam easy hai — bas signs ke saath jod do, koi component-vomponent nahi. > > Point charge ka potential nikalne ke liye hum field $\vec E = kq/r^2$ ko infinity se $r$ tak integrate karte hain (minus ke saath, kyunki agent field ke against push karta hai). Result milta hai $V = kq/r$. Notice karo: field me $r^2$ neeche hai, potential me sirf $r$ — isliye potential dheere girta hai, field jaldi. Aur ulta jaana ho to $\vec E = -\nabla V$, yaani field potential ki **slope** (dhalaan) hai, minus sign ke saath — field hamesha "neeche" yaani lower potential ki taraf point karta hai. > > Best mental picture: $V$ ek pahaadi (hill) ki **height** hai, aur $\vec E$ wo push hai jo ball ko sabse steep dhalaan ke neeche dhakelti hai. Flat zameen = no field. Zyada steep = strong field. Isiliye ek important trick yaad rakho: jahan $\vec E=0$ ho zaroori nahi $V=0$ ho (jaise do same positive charges ke beech midpoint par field cancel par potential strong). Exam me yahi log galti karte hain — field ko scalar ki tarah add kar dete hain. Field vector hai (components se add), potential scalar hai (signs ke saath direct add). Isko pakka rakhoge to ye topic full marks ka hai. ![[audio/1.8.09-Potential-of-point-charge,-potential-from-field-and-vice-versa.mp3]]