Exercises — Equipotential surfaces — perpendicular to field

WHAT: We test the parallel-vs-perpendicular claim. WHY: From $dV=-\vec{E}\cdot d\vec{l}=0$ for in-surface steps, $\vec{E}$ is normal (perpendicular) to the surface. Answer: FALSE. Field lines cross equipotentials at $==90^{\circ }==$, never parallel. Mantra: E cuts V at 90.

WHAT: Set $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}=\text{const}$. WHY: With $Q$ fixed, $V$ constant forces $r$ constant — one fixed distance from the charge. What it looks like: all points at fixed distance = a sphere. So equipotentials are concentric spheres centred on $Q$ (see the picture below). The field $\vec{E}$ points radially, along the outward normal $\hat{r}$, so it pierces each sphere at $90^{\circ }$. ✓

WHAT: Use $W_{AB}=q(V_A-V_B)$. WHY: Same equipotential means $V_A=V_B$, so the bracket is zero. $W_{AB}=q(V_A-V_B)=q\cdot 0=\boxed{0\text{J}}$ True for any path on the surface — the electric force is conservative (see Work and Conservative Forces).

WHAT: Use $E=\left|\frac{dV}{dn}\right|$ — magnitude of potential change per perpendicular metre. WHY this tool: $E=-\frac{dV}{dn}$ links spacing of equipotentials to field strength; we want a magnitude so we drop the sign. $E=\frac{\Delta V}{\Delta n}=\frac{12\text{V}}{4\times 10^{-3}\text{m}}=\boxed{3000\text{V/m}}$ What it looks like: two parallel chalk lines $4\text{mm}$ apart; the field arrows run straight across them from the higher-$V$ plane to the lower-$V$ one. Compare with the Parallel Plate Capacitor.

WHAT: Invert $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$ for $r$. WHY: The equipotential of value $V$ is the sphere of that radius; solving for $r$ locates it. $r=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{V}=\frac{(8.99\times 10^9)(2.0\times 10^{-9})}{180}=\boxed{0.0999\text{m}\approx 10\text{cm}}$

WHAT: Use the 1-D gradient $E_x=-\frac{dV}{dx}$. WHY this tool: $\vec{E}=-\nabla V$; in one dimension the gradient is just the ordinary derivative (see Gradient and Directional Derivative). $\frac{dV}{dx}=-500\text{V/m}\Rightarrow E_x=-(-500)=+500\text{V/m}$ So $\vec{E}=+500\hat{x}\text{V/m}$, pointing toward $\boxed{+x}$ — i.e. toward lower potential (V decreases as $x$ grows). ✓

WHAT: Perpendicular spacing $\Delta n\approx \frac{\Delta V}{E}$ for a chosen $\Delta V$. WHY: $E=\frac{dV}{dn}\Rightarrow \Delta n=\frac{\Delta V}{E}$; bigger $E$ packs equipotentials closer. At $x=1$: $E=\frac{90}{1^2}=90\text{V/m}$, so $\Delta n=\frac{30}{90}=\boxed{0.333\text{m}}$. At $x=3$: $E=\frac{90}{3^2}=10\text{V/m}$, so $\Delta n=\frac{30}{10}=\boxed{3\text{m}}$. Pattern & what it looks like: near the charge (small $x$) equipotentials bunch up (strong field); far away they spread out ninefold (weak field). This is the map-contour intuition — cliffs vs gentle slopes.

WHAT: $r=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{V}$ for each value. $r_1=\frac{(8.99\times 10^9)(1.0\times 10^{-9})}{100}=\boxed{0.0899\text{m}},r_2=\frac{(8.99\times 10^9)(1.0\times 10^{-9})}{50}=\boxed{0.1798\text{m}}$ WHY no crossing: each point in space has exactly one distance $r$ from the charge, hence exactly one value of $V$. A crossing point would need two potentials ($100$ and $50\text{V}$) at once — impossible. The two spheres are nested, never touching.

WHAT: On any equipotential, $\vec{E}$ is perpendicular to the surface, so the angle with the surface is fixed by geometry, not by the numbers. WHY: From $dV=-\vec{E}\cdot d{\vec{l}}_{\text{tangent}}=0$, the field is normal to the surface. Hence the angle between $\vec{E}$ and the surface is $\boxed{90^{\circ }}$. Local tangent: the surface tangent is the direction $\perp$ to $\vec{E}$. $\vec{E}$ has direction angle $\arctan (4/3)=53.13^{\circ }$ from $+x$, so the tangent line runs at $53.13^{\circ }+90^{\circ }=143.13^{\circ }$ (equivalently $-36.87^{\circ }$). The magnitude $|\vec{E}|=\sqrt{3^2+4^2}\times 10^3=5000\text{V/m}$ is the local field strength (see Conductors in Electrostatic Equilibrium).

(a) WHAT: Uniform field ⇒ $\Delta V=Ed$. WHY: For constant $E$, $V$ is linear across the gap; $|\Delta V|=E\cdot (\text{perpendicular gap})$. $\Delta V=Ed=(2.0\times 10^4)(5\times 10^{-3})=\boxed{100\text{V}}$ (b) WHAT: Work depends only on endpoints: $W=q(V_A-V_B)=q\Delta V$. WHY: The sideways part lies along equipotential planes (contributing zero work); only the perpendicular crossing of $100\text{V}$ matters. The $8\text{mm}$ diagonal length is a distractor. $W=q\Delta V=(2\times 10^{-6})(100)=\boxed{2\times 10^{-4}\text{J}}$ The field does positive work pushing the positive charge from high $V$ to low $V$. ✓

(a) WHAT: $\vec{E}=-\nabla V=-\left(\frac{\partial V}{\partial x}\hat{x}+\frac{\partial V}{\partial y}\hat{y}\right)$. $\frac{\partial V}{\partial x}=6,\frac{\partial V}{\partial y}=-8\Rightarrow \vec{E}=-(6\hat{x}-8\hat{y})=\boxed{-6\hat{x}+8\hat{y}}\text{V/m}$ (b) Equipotentials: $6x-8y=\text{const}\Rightarrow y=\frac{6}{8}x-\frac{\text{const}}{8}$, a straight line of slope $\frac{6}{8}=\boxed{0.75}$. Perpendicular check: $\vec{E}$ has slope $\frac{E_y}{E_x}=\frac{8}{-6}=-\frac{4}{3}$. Product of slopes $=0.75\times (-\frac{4}{3})=-1$. Two lines whose slopes multiply to $-1$ are perpendicular. ✓ So $\vec{E}\perp$ equipotential, exactly as promised.

WHAT: Take any displacement $d\vec{l}$ that stays on the surface (a tangent step). WHY: Tangent steps are exactly the ones that keep $V$ fixed. Since $V$ is constant on the surface, $dV=0$ for every such tangent step: $0=dV=\nabla V\cdot d{\vec{l}}_{\text{tangent}}.$ A dot product of two nonzero vectors is zero only when they are perpendicular. This holds for every tangent direction $d\vec{l}$, so $\nabla V$ is perpendicular to all of them — i.e. normal to the surface. Then $\vec{E}=-\nabla V\Rightarrow \vec{E}\parallel \nabla V\Rightarrow \vec{E}\text{is also normal to the surface.}■$ The minus sign only flips $\vec{E}$ to point toward decreasing $V$; it does not change the perpendicularity.

WHAT: $r=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{V}$ for each $V$. With $\frac{1}{4\pi {\epsilon }_0}Q=(8.99\times 10^9)(4.0\times 10^{-9})=35.96\text{V⋅m}$: $r_{400}=\frac{35.96}{400}=\boxed{0.0899\text{m}},r_{300}=\frac{35.96}{300}=\boxed{0.1199\text{m}}$ $r_{200}=\frac{35.96}{200}=\boxed{0.1798\text{m}},r_{100}=\frac{35.96}{100}=\boxed{0.3596\text{m}}$ Spacings: $r_{300}-r_{400}=0.0300\text{m}$; $r_{200}-r_{300}=0.0599\text{m}$; $r_{100}-r_{200}=0.1798\text{m}$. Each gap is larger than the last (equal $\Delta V$ but wider spacing), so $E=\Delta V/\Delta n$ shrinks outward — matching $E\propto 1/r^2$. ✓ (See the widening rings in the figure below.)

WHAT / WHY: With $\vec{E}=-\nabla V=0$, all partial derivatives of $V$ vanish, so $V$ does not change anywhere inside the metal — it is one single value. The whole conductor (surface included) is one equipotential. The perpendicular statement in the zero case: "$\vec{E}\perp$ surface" is vacuously true — a zero vector has no direction to violate perpendicularity, and there is no in-surface component to cancel. The surface just outside still has $\vec{E}$ perpendicular to it, because any tangential piece would drive the free charges until it cancels (see Conductors in Electrostatic Equilibrium). ✓