Question bank — Electric field — definition, field lines, superposition

False. Ye $\vec{E}$ ke along accelerate karta hai, lekin iske paas inertia aur pehle se velocity bhi hai — iska path line ke saath tabhi hoga jab ye rest se shuru ho; warna ye lines ke across drift karega jaise ek ball gravity mein sideways pheki jaati hai.

True. Definition ke hisaab se $\vec{E}$ force-per-positive-charge hai, aur ek line ki tangent $\vec{E}$ hai, isliye ek positive charge ko line ke arrow ke saath push kiya jaata hai.

False. $\vec{E}$ empty space mein kisi point par zero ho sakta hai — jaise do equal positive charges ke beech mein, jahan do fields cancel ho jaati hain. Zero field ≠ zero nearby charge.

False. $\vec{E}=\vec{F}/q_0$: force bhi double hoti hai, isliye ratio unchanged rehta hai. $q_0$ se divide karne ka poora point yahi hai — field sources ki property hai, probe ki nahi.

False. Touch karne ka matlab hai ek point share karna jahan same tangent direction ho; matlab $\vec{E}$ ka wahan ek hi value hona — jo theek hai — lekin phir wo coincide karte hain, do alag lines nahi rehte. Asli crossing (ek point par do directions) forbidden hai.

True. $\vec{E}=kq/r^2$ jaise $r\to 0$ hota hai, infinity mein blow up karta hai; formula infinity deta hai, isliye point charge ki apni location ek singularity hai (ek idealization jo real spread-out charge avoid karti hai).

False. Lines negative charges mein enter karti hain (unki taraf point karti hain). Ek positive test charge negative source ke paas andar ki taraf kheencha jaata hai, isliye $\vec{E}$ — aur line ka arrow — negative charge ki taraf point karta hai.

False. Aap jitni total lines draw karte ho wo sirf ek convention hai; jo strength encode karta hai wo line density hai (lines per unit area crossed). Charge double karne se lines double hoti hain aur wo patli bhi ho sakti hain, isliye sirf bada count akela stronger field guarantee nahi karta.

Error: ye vectors hain. Tumhe har ek ko $x,y$ (aur $z$) components mein resolve karke componentwise add karna hoga. Magnitudes tabhi add hoti hain jab teeno fields parallel hon.

Error: $\hat{r}$ source se field point ki taraf point karta hai. Phir $q$ ka sign direction handle karta hai: positive $q$ $\vec{E}$ ko $\hat{r}$ ke saath deta hai (outward), negative $q$ ise inward flip karta hai.

Error: ek charge apne aap par koi net force exert nahi karta. Kisi point par $\vec{E}$ sirf doosre (source) charges se build hota hai jo us location par evaluate kiye jaate hain.

Error: ye ulta hai — axes aise set karo ki charges x-axis par hon (figure s01, right panel dekho). Do y-components ke opposite signs hote hain aur cancel hote hain; do x-components dono $-x$ direction mein point karte hain aur add hote hain. Dekho Electric Dipole.

Error: sirf tabhi jab net charge nonzero ho. Ek dipole (zero net charge) $1/r^3$ ki tarah fall karta hai; ek quadrupole aur bhi tezi se. Leading term woh lowest nonzero multipole hai.

Error: field positive test charge par push se define hoti hai, jo repel hota hai — isliye positive source ka $\vec{E}$ outward point karta hai. Opposites ka attraction negative test charge se aata hai, source ki field se nahi.

Error: tum vector fields ${\vec{E}}_i$ add karte ho, jo apne apne $r_i$ se already computed hain. Tum kabhi denominators combine nahi karte — har source apni distance aur direction ${\hat{r}}_i$ rakhta hai. Ye Coulomb's Law hai jo har source par ek baar apply hota hai, phir sum kiya jaata hai.

Taaki field sirf source aur location describe kare, chahe hum koi bhi probe use karein. Divide karne se probe answer se remove ho jaata hai, $\vec{E}$ ko space ki ek property bana deta hai.

Ek real test charge sources ko wapas push karta hai aur unhe relocate karta hai, woh field badal deta hai jo hum measure karna chahte hain. Limit ensure karti hai ki probe bina disturb kiye read kare.

Ek point charge se same fixed number of lines ek sphere ke area $4\pi r^2$ par spread hoti hain. Density = lines/area $\propto 1/r^2$ — geometry hi Coulomb's Law ka $1/r^2$ falloff hai.

Kyunki Coulomb's law charge mein linear hai aur forces vectors ki tarah add hote hain (Newton). Linearity ka matlab hai ki har source aise act karta hai jaise doosre wahan hain hi nahi, aur results sirf sum hote hain — Superposition Principle ka essence yahi hai.

Door se $+q$ aur $-q$ almost cancel ho jaate hain; sirf unke separation se thoda sa leftover bachta hai, aur woh residue $r$ ki ek aur power tezi se shrink karta hai. Dekho Electric Dipole.

Charge B ke baare mein door ki charge A ko mysteriously "jaanne" ki jagah, A space ko field se bharta hai aur B sirf wahan ke field ko respond karta hai jahan B baitha hai — koi gap ke paas action nahi.

Lines ek continuous vector field ki sampling hain. Do lines ke beech field abhi bhi exist karti hai; hum readability ke liye finite kuch draw karte hain, strength encode karne ke liye density choose karte hain.

Zero. Do fields magnitude mein equal aur direction mein opposite hain (figure s01, left), isliye cancel ho jaati hain — wahan ek positive test charge koi push feel nahi karta (ek unstable equilibrium).

Nahi. Dono fields same direction mein point karti hain ($+$ se $-$ ki taraf, figure s01 right) aur add hoti hain. Midpoint fields sirf tabhi oppose aur cancel hoti hain jab charges same sign ke hon.

Ye $1/r^2$ ki tarah zero ki taraf tend karta hai, lekin kabhi exactly zero nahi pahunchta — field ki infinite range hoti hai, sirf door jaane par vanishingly weak ho jaati hai.

Haan — us point par do source contributions ek single resultant vector mein superpose hote hain jiska ek hi direction hota hai. Apparent "meeting" sirf woh hai jahan ek net line guzarti hai; field perfectly single-valued hai.

Koi nahi (usse $\vec{E}=0$ sab jagah). Koi charge nahi toh $kq/r^2$ mein koi source term nahi; wo na field lines emit karta hai na terminate.

Nahi — field sirf weak hai, absent nahi. Low line density small $|\vec{E}|$ signal karta hai; ek line principle mein har point se guzarti hai.

Andar sab jagah Zero. Charges conductor ki surface par rearrange hote hain jab tak interior field completely cancel na ho jaaye — electrostatic equilibrium ka ek classic result (provable with Gauss's Law). Isliye ek metal box apna interior shield karta hai.

Surface ke perpendicular. Koi bhi tangential component surface charges ko sideways push karta jab tak wo rukna band na kar dein, isliye equilibrium par sirf perpendicular part bachta hai.