Exercises — Electric field of point charge, dipole, ring, disk, line charge (Gauss's law)

Parent ke summary table se seedha padho.

• (a) Point charge: field lines area $4\pi r^2$ wale sphere par spread hoti hain, isliye density $\propto 1/r^2$ → $n=2$.
• (b) Infinite line: lines area $2\pi rL$ wale cylinder par spread hoti hain, isliye $\propto 1/r$ → $n=1$.
• (c) Infinite sheet: lines parallel rehti hain, kabhi spread nahi hoti → $n=0$ (constant field).
• (d) Dipole: do opposite $1/r^2$ terms almost cancel ho jaate hain, bacha rehta hai $n=3$. Mnemonic: Point Two, Dipole Three, Line One, Sheet None.

• Ring: koi aisi Gaussian surface exist nahi karti jis par $E$ dono constant ho aur perpendicular bhi ho (ring ki symmetry spherical, cylindrical, ya planar nahi hai). → integrate karo elements $dq$ par.
• Infinite cylinder: poori cylindrical symmetry hai → coaxial cylinder wrap karo, $E$ flux integral se bahar aa jaata hai → Gauss's law. Deciding question hamesha yahi hai: "Kya main aisi surface draw kar sakta hoon jahan $\vec{E}\cdot d\vec{A}$ bas $E\times A$ ban jaaye?" Agar haan → Gauss. Agar nahi → integrate karo.

KYA: $E=kq/r^2$ mein plug in karo. YEH formula kyun: ek akela point charge, radial symmetry. $E=\frac{(8.99\times 10^9)(2.0\times 10^{-9})}{(0.30{)}^2}=\frac{17.98}{0.09}\approx 2.0\times 10^2\text{N/C}.$ $E\approx 200\text{N/C}$, radially outward point kar raha hai (kyunki $q>0$).

SIRF axial component kyun bachta hai: har element $dq$ ke liye, diametrically opposite element uski sideways pull cancel kar deta hai; sirf along-axis piece (weighted by $\cos \theta =x/\sqrt{R^2+x^2}$) bachti hai — figure dekho.

Gauss kyun: cylindrical symmetry. Parent se, $E=\frac{2k\lambda }{r}$. $E=\frac{2(8.99\times 10^9)(3.0\times 10^{-6})}{0.20}=\frac{5.394\times 10^4}{0.20}\approx 2.70\times 10^5\text{N/C},$ wire se radially door point kar raha hai.

KYA: $E(x)=kQx(R^2+x^2{)}^{-3/2}$ maximise karo. Differentiate kyun: peak wahan hai jahan slope $dE/dx=0$ ho. $\frac{dE}{dx}=kQ\left[(R^2+x^2{)}^{-3/2}+x\cdot (-\frac{3}{2})(R^2+x^2{)}^{-5/2}(2x)\right].$ $(R^2+x^2{)}^{-5/2}$ factor out karo: $\frac{dE}{dx}=kQ(R^2+x^2{)}^{-5/2}[(R^2+x^2)-3x^2]=kQ(R^2+x^2{)}^{-5/2}(R^2-2x^2).$ Bracket zero karo: $R^2-2x^2=0\Rightarrow x=R/\sqrt{2}$. ✓ Peak value: $x^2=R^2/2$ ke saath, $R^2+x^2=\frac{3}{2}{R}^2$: $E_{max}=\frac{kQ(R/\sqrt{2})}{(\frac{3}{2}{R}^2{)}^{3/2}}=\frac{kQ}{R^2}\cdot \frac{1/\sqrt{2}}{(3/2{)}^{3/2}}=\frac{2kQ}{3\sqrt{3}{R}^2}\approx \frac{0.385kQ}{R^2}.$

(a) $x\ll R$: tab $\frac{x}{\sqrt{R^2+x^2}}\to 0$, bracket $\to 1$: $E\to \frac{\sigma }{2{\epsilon }_0}(\text{constant — infinite sheet}).$ (b) $x\gg R$: likho $\frac{x}{\sqrt{R^2+x^2}}=\frac{1}{\sqrt{1+R^2/x^2}}\approx 1-\frac{R^2}{2x^2}$ (binomial, kyunki $R^2/x^2$ tiny hai). Bracket $=1-\left(1-\frac{R^2}{2x^2}\right)=\frac{R^2}{2x^2}$. Toh $E\to \frac{\sigma }{2{\epsilon }_0}\cdot \frac{R^2}{2x^2}=\frac{\sigma \pi R^2}{4\pi {\epsilon }_0{x}^2}=\frac{kQ}{x^2},$ total charge $Q=\sigma \pi R^2$ use karke. Door se ek disk point charge jaisi dikhti hai. ✓

Parent se: $E_{ax}=\frac{2kp}{r^3}$ ($+\vec{p}$ ke along) aur $E_{eq}=\frac{kp}{r^3}$ ($-\vec{p}$ ke along). $\frac{E_{ax}}{E_{eq}}=\frac{2kp/r^3}{kp/r^3}=2.$ Directions: axial field $\vec{p}$ ke along point karta hai ($-$ se $+$ ki taraf aur aage); equatorial field $\vec{p}$ ke opposite point karta hai. Dekho Electric dipole in a uniform field ki aisa dipole phir external field mein kaise behave karta hai.

Superposition kyun: total field = har sheet ke field ka vector sum, har ek ki magnitude $\frac{\sigma }{2{\epsilon }_0}$ — $+$ se door aur $-$ ki taraf.

• Beech mein: dono fields same direction mein point karte hain ($+$ plate se $-$ plate ki taraf), isliye add ho jaate hain: $E=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}.$
• Bahar: dono fields opposite directions mein point karte hain aur cancel ho jaate hain: $E=0$. Yahi exactly Parallel plate capacitor result hai: andar uniform $\sigma /{\epsilon }_0$, bahar zero.

KYA: centre par, har element same distance $R$ par hai, isliye har element contribute karta hai $dE=\frac{kdq}{R^2}$ jo element se centre ki taraf point karta hai ($Q>0$ ke liye, actually door — direction carefully track karte hain). Setup: semicircle angle $\varphi$ mein $-90^{\circ }$ se $+90^{\circ }$ tak span karta hai $+y$ ki taraf opening ke saath. Element charge $dq=\frac{Q}{\pi R}Rd\varphi =\frac{Q}{\pi }d\varphi$ (kyunki linear density $\lambda =Q/(\pi R)$, arc length $Rd\varphi$). Symmetry se $x$-components cancel ho jaate hain; sirf symmetry axis ke along component bachta hai. Centre par har element ka field uski apni radius ke along radially inward-to-outward point karta hai; jo component bachta hai woh axis se measure karke $dE\cos \varphi$ hai. $E={\int }_{-\pi /2}^{\pi /2}\frac{kdq}{R^2}\cos \varphi =\frac{k}{R^2}\cdot \frac{Q}{\pi }{\int }_{-\pi /2}^{\pi /2}\cos \varphi d\varphi =\frac{kQ}{\pi R^2}[\sin \varphi {]}_{-\pi /2}^{\pi /2}.$ $E=\frac{kQ}{\pi R^2}(1-(-1))=\frac{2kQ}{\pi R^2}.$ Direction: symmetry axis ke along, charged arc se door point karta hai ($Q>0$ ke liye).

(a) $\frac{\sigma }{2{\epsilon }_0}=\frac{4.0\times 10^{-6}}{2(8.85\times 10^{-12})}=2.26\times 10^5\text{N/C}$. Bracket: $1-\frac{0.20}{\sqrt{0.20^2+0.20^2}}=1-\frac{0.20}{0.2828}=1-0.7071=0.2929$. $E=2.26\times 10^5\times 0.2929\approx 6.62\times 10^4\text{N/C}.$ (b) Infinite sheet deta $2.26\times 10^5\text{N/C}$. $x=R$ par hume sirf $\approx 29\%$ milta hai. Kyun: $R$ ke comparable distance par, disk ab "tumhara poora aasman nahi bharta" — tum uska edge dekhte ho, isliye kam charge tumhare saamne hai, aur field idealized infinite sheet se bahut kamzor hai.

Yahan $d=2.0\text{mm}=0.002\text{m}$, $p=qd=(1.0\times 10^{-9})(0.002)=2.0\times 10^{-12}\text{C⋅m}$. Direct Coulomb at $r=0.50$ axis par: $+q$ distance $r-d/2=0.499$ par hai, $-q$ distance $r+d/2=0.501$ par. $E=kq\left[\frac{1}{0.499^2}-\frac{1}{0.501^2}\right]=(8.99)(1.0\times 10^{-9}\cdot 10^9)\text{scaled}.$ Carefully compute karo: $kq=(8.99\times 10^9)(1.0\times 10^{-9})=8.99$. $E=8.99\left[\frac{1}{0.249001}-\frac{1}{0.251001}\right]=8.99[4.01605-3.98405]=8.99(0.03200)\approx 0.2877\text{N/C}.$ Formula check: $\frac{2kp}{r^3}=\frac{2(8.99\times 10^9)(2.0\times 10^{-12})}{0.50^3}=\frac{0.03596}{0.125}\approx 0.2877\text{N/C}$. ✓ Dono agree karte hain.

$E_x=-dV/dx$ kyun: field woh (minus) hai jitni steeply potential space mein drop hoti hai — $V$ ka "downhill slope." $V(x)=kQ(R^2+x^2{)}^{-1/2}.$ $\frac{dV}{dx}=kQ\cdot (-\frac{1}{2})(R^2+x^2{)}^{-3/2}(2x)=-\frac{kQx}{(R^2+x^2{)}^{3/2}}.$ $E_x=-\frac{dV}{dx}=\frac{kQx}{(R^2+x^2{)}^{3/2}}.✓$ Exactly parent ka ring formula — do raaste, ek field.

Kaunse source ka field distance par depend nahi karta? ::: Infinite sheet ka, $E=\sigma /2{\epsilon }_0$. Ring ke axis par $E$ sabse bada kahan hota hai? ::: $x=R/\sqrt{2}$ par. Equal $r$ par axial aur equatorial dipole field ka ratio? ::: $2:1$. $\sigma$ density ke do opposite sheets ke beech field? ::: $\sigma /{\epsilon }_0$ (bahar: $0$). Axis par field aur potential ka relation? ::: $E_x=-dV/dx$.