Exercises — Equipotential surfaces — perpendicular to field

WHAT: Hum parallel-vs-perpendicular claim test kar rahe hain. WHY: $dV=-\vec{E}\cdot d\vec{l}=0$ se, jo in-surface steps ke liye hota hai, $\vec{E}$ surface ke normal (perpendicular) hota hai. Answer: GALAT. Field lines equipotentials ko $==90^{\circ }==$ par cross karti hain, kabhi parallel nahi. Mantra: E cuts V at 90.

WHAT: $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}=\text{const}$ set karo. WHY: $Q$ fixed hone ke saath, $V$ constant hona $r$ ko constant force karta hai — charge se ek fixed distance. Kaisa dikhta hai: ek fixed distance par saare points = ek sphere. Toh equipotentials concentric spheres hain jo $Q$ par centred hain (neeche picture dekho). Field $\vec{E}$ radially point karta hai, outward normal $\hat{r}$ ke along, toh yeh har sphere ko $90^{\circ }$ par pierce karta hai. ✓

WHAT: $W_{AB}=q(V_A-V_B)$ use karo. WHY: Ek hi equipotential ka matlab hai $V_A=V_B$, toh bracket zero hai. $W_{AB}=q(V_A-V_B)=q\cdot 0=\boxed{0\text{J}}$ Surface par kisi bhi path ke liye sahi — electric force conservative hai (dekho Work and Conservative Forces).

WHAT: $E=\left|\frac{dV}{dn}\right|$ use karo — perpendicular metre per potential change ka magnitude. WHY this tool: $E=-\frac{dV}{dn}$ equipotentials ki spacing ko field strength se link karta hai; hum magnitude chahte hain toh sign drop karte hain. $E=\frac{\Delta V}{\Delta n}=\frac{12\text{V}}{4\times 10^{-3}\text{m}}=\boxed{3000\text{V/m}}$ Kaisa dikhta hai: do parallel chalk lines $4\text{mm}$ apart; field arrows unke across seedhe chalte hain higher-$V$ plane se lower-$V$ plane ki taraf. Parallel Plate Capacitor se compare karo.

WHAT: $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$ ko $r$ ke liye invert karo. WHY: Value $V$ ka equipotential us radius ki sphere hai; $r$ ke liye solve karne se uski location milti hai. $r=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{V}=\frac{(8.99\times 10^9)(2.0\times 10^{-9})}{180}=\boxed{0.0999\text{m}\approx 10\text{cm}}$

WHAT: 1-D gradient $E_x=-\frac{dV}{dx}$ use karo. WHY this tool: $\vec{E}=-\nabla V$; ek dimension mein gradient sirf ordinary derivative hai (dekho Gradient and Directional Derivative). $\frac{dV}{dx}=-500\text{V/m}\Rightarrow E_x=-(-500)=+500\text{V/m}$ Toh $\vec{E}=+500\hat{x}\text{V/m}$, $\boxed{+x}$ ki taraf point karta hai — yaani lower potential ki taraf ($V$ kam hota hai jaise $x$ badhta hai). ✓

WHAT: Perpendicular spacing $\Delta n\approx \frac{\Delta V}{E}$ ek chosen $\Delta V$ ke liye. WHY: $E=\frac{dV}{dn}\Rightarrow \Delta n=\frac{\Delta V}{E}$; bada $E$ equipotentials ko closer pack karta hai. $x=1$ par: $E=\frac{90}{1^2}=90\text{V/m}$, toh $\Delta n=\frac{30}{90}=\boxed{0.333\text{m}}$. $x=3$ par: $E=\frac{90}{3^2}=10\text{V/m}$, toh $\Delta n=\frac{30}{10}=\boxed{3\text{m}}$. Pattern aur kaisa dikhta hai: charge ke paas (chota $x$) equipotentials bunch up ho jaate hain (strong field); door par woh naunguna spread out ho jaate hain (weak field). Yeh map-contour intuition hai — cliffs vs gentle slopes.

WHAT: Har value ke liye $r=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{V}$. $r_1=\frac{(8.99\times 10^9)(1.0\times 10^{-9})}{100}=\boxed{0.0899\text{m}},r_2=\frac{(8.99\times 10^9)(1.0\times 10^{-9})}{50}=\boxed{0.1798\text{m}}$ Kyun koi crossing nahi: space mein har point ka charge se exactly ek distance $r$ hota hai, isliye exactly ek value $V$ hoti hai. Ek crossing point ko ek saath do potentials ($100$ aur $50\text{V}$) chahiye honge — impossible. Dono spheres nested hain, kabhi touch nahi karti.

WHAT: Kisi bhi equipotential par, $\vec{E}$ surface ke perpendicular hota hai, toh surface ke saath angle geometry se fixed hai, numbers se nahi. WHY: $dV=-\vec{E}\cdot d{\vec{l}}_{\text{tangent}}=0$ se, field surface ke normal hota hai. Isliye $\vec{E}$ aur surface ke beech angle $\boxed{90^{\circ }}$ hai. Local tangent: surface tangent woh direction hai jo $\vec{E}$ ke $\perp$ hai. $\vec{E}$ ka direction angle $\arctan (4/3)=53.13^{\circ }$ hai $+x$ se, toh tangent line $53.13^{\circ }+90^{\circ }=143.13^{\circ }$ par chalti hai (equivalently $-36.87^{\circ }$). Magnitude $|\vec{E}|=\sqrt{3^2+4^2}\times 10^3=5000\text{V/m}$ local field strength hai (dekho Conductors in Electrostatic Equilibrium).

(a) WHAT: Uniform field ⇒ $\Delta V=Ed$. WHY: Constant $E$ ke liye, $V$ gap ke across linear hai; $|\Delta V|=E\cdot (\text{perpendicular gap})$. $\Delta V=Ed=(2.0\times 10^4)(5\times 10^{-3})=\boxed{100\text{V}}$ (b) WHAT: Kaam sirf endpoints par depend karta hai: $W=q(V_A-V_B)=q\Delta V$. WHY: Sideways part equipotential planes ke along chalti hai (zero work contribute karta hai); sirf $100\text{V}$ ka perpendicular crossing matter karta hai. $8\text{mm}$ diagonal length ek distractor hai. $W=q\Delta V=(2\times 10^{-6})(100)=\boxed{2\times 10^{-4}\text{J}}$ Field positive kaam karta hai positive charge ko high $V$ se low $V$ ki taraf push karke. ✓

(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: $6x-8y=\text{const}\Rightarrow y=\frac{6}{8}x-\frac{\text{const}}{8}$, slope $\frac{6}{8}=\boxed{0.75}$ ki ek straight line. Perpendicular check: $\vec{E}$ ka slope $\frac{E_y}{E_x}=\frac{8}{-6}=-\frac{4}{3}$ hai. Slopes ka product $=0.75\times (-\frac{4}{3})=-1$. Jin do lines ke slopes $-1$ se multiply hote hain woh perpendicular hain. ✓ Toh $\vec{E}\perp$ equipotential, bilkul jaisa promise kiya gaya tha.

WHAT: Koi bhi displacement $d\vec{l}$ lo jo surface par remain kare (ek tangent step). WHY: Tangent steps exactly wahi hain jo $V$ ko fixed rakhte hain. Kyunki $V$ surface par constant hai, $dV=0$ har aise tangent step ke liye: $0=dV=\nabla V\cdot d{\vec{l}}_{\text{tangent}}.$ Do nonzero vectors ka dot product zero tab hi hota hai jab woh perpendicular hon. Yeh har tangent direction $d\vec{l}$ ke liye hold karta hai, toh $\nabla V$ unme se sabhi ke perpendicular hai — yaani surface ke normal. Phir $\vec{E}=-\nabla V\Rightarrow \vec{E}\parallel \nabla V\Rightarrow \vec{E}\text{surface ke bhi normal hai.}■$ Minus sign sirf $\vec{E}$ ko decreasing $V$ ki taraf flip karta hai; yeh perpendicularity change nahi karta.

WHAT: Har $V$ ke liye $r=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{V}$. $\frac{1}{4\pi {\epsilon }_0}Q=(8.99\times 10^9)(4.0\times 10^{-9})=35.96\text{V⋅m}$ ke saath: $r_{400}=\frac{35.96}{400}=\boxed{0.0899\text{m}},r_{300}=\frac{35.96}{300}=\boxed{0.1199\text{m}}$ $r_{200}=\frac{35.96}{200}=\boxed{0.1798\text{m}},r_{100}=\frac{35.96}{100}=\boxed{0.3596\text{m}}$ Spacings: $r_{300}-r_{400}=0.0300\text{m}$; $r_{200}-r_{300}=0.0599\text{m}$; $r_{100}-r_{200}=0.1798\text{m}$. Har gap pichle se bada hai (equal $\Delta V$ lekin wider spacing), toh $E=\Delta V/\Delta n$ outward shrink karta hai — $E\propto 1/r^2$ se match karta hai. ✓ (Neeche figure mein widening rings dekho.)

WHAT / WHY: $\vec{E}=-\nabla V=0$ ke saath, $V$ ke saare partial derivatives vanish ho jaate hain, toh $V$ metal ke andar kahin bhi change nahi hota — yeh ek single value hai. Poora conductor (surface sameta) ek equipotential hai. Zero case mein perpendicular statement: "$\vec{E}\perp$ surface" vacuously true hai — ek zero vector ki koi direction nahi hoti jo perpendicularity violate kare, aur koi in-surface component nahi hai jo cancel ho. Surface ke just bahar ka $\vec{E}$ abhi bhi perpendicular hai uske, kyunki koi bhi tangential piece free charges ko drive kar deta jab tak woh cancel na ho jaaye (dekho Conductors in Electrostatic Equilibrium). ✓