Coulomb's law — force, comparison with gravity: Exercises

1.8.2 · D4 · Physics › Electromagnetism › Coulomb's law — force, comparison with gravity

Poore document mein, hum wohi constant use karte hain jo parent note ne establish kiya: $k = \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}$ aur magnitude form $F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}} .$

Ek notation jo hum baar baar use karte hain: jab do charges ko 1 aur 2 label kiya jaaye, toh hum $F_{12}$ likhte hain jiska matlab hai "charge 1 par charge 2 ki wajah se force" (pehla subscript woh charge hai jo force feel karta hai, doosra woh charge hai jo force cause karta hai). Toh $F_{21}$ iska ulta hai: charge 2 par force jo charge 1 ki wajah se lagti hai. Newton's third law kehta hai ki yeh dono size mein equal aur direction mein opposite hain, $F_{12} = - F_{21}$ .

Yaad rakhne wali baatein jo har baar kaam aayengi:

$μ C$ ka matlab hai $\times 1 0^{- 6} C$ , $n C$ ka matlab hai $\times 1 0^{- 9} C$ .
Distances metres mein daalni hain, charges coulombs mein — sirf SI units.
Scalar formula mein magnitudes use karo; attract/repel ka faisla signs se alag karo.

Level 1 — Recognition

L1.1

In mein se kaun se pairs repel karte hain, aur kaun se attract? (a) $+ 4 μ C$ aur $+ 1 μ C$ (b) $- 3 μ C$ aur $+ 2 μ C$ (c) $- 5 μ C$ aur $- 5 μ C$ .

Recall Solution

Rule yeh hai: like charges repel karte hain, unlike charges attract karte hain. Signs ka product lo.

(a) $(+) (+) = +$ → repel.
(b) $(-) (+) = -$ → attract.
(c) $(-) (-) = +$ → repel. Gaur karo ki humne formula chhua hi nahi — sirf product ka sign direction decide karta hai.

L1.2

$+ 2 μ C$ aur $+ 8 μ C$ ka charge ek fixed distance par baithe hain. Bina kisi calculation ke, kaun sa charge zyada force feel karta hai?

Recall Solution

Koi nahi — dono forces ki magnitude equal hai. Hamara subscript convention yaad karo: $F_{12}$ charge 1 par charge 2 ki wajah se force hai, aur $F_{21}$ charge 2 par charge 1 ki wajah se force hai. Newton's third law se, $F_{12} = - F_{21}$ : charge 1 par charge 2 ka push exactly utna hi bada hai jitna charge 2 par charge 1 ka push. Formula $F = k ∣ q_{1} q_{2} ∣/ r^{2}$ product $q_{1} q_{2}$ use karta hai, jo same rehta hai chahe tum kisi bhi charge ko "force feel karne wala" kaho. Ek bada charge aur ek chhota charge equal size ka push feel karte hain — chhota wala sirf zyada accelerate karta hai (yeh mass ki baat hai, force ki nahi).

Level 2 — Application

L2.1

Do charges $q_{1} = + 5 μ C$ aur $q_{2} = + 2 μ C$ ek doosre se $0.20 m$ door hain. Force ki magnitude nikalo.

Recall Solution

Step 1 — SI units. $q_{1} = 5 \times 1 0^{- 6} C$ , $q_{2} = 2 \times 1 0^{- 6} C$ , $r = 0.20 m$ . Kyun: formula coulombs aur metres ke liye calibrated hai.

Step 2 — magnitude form mein plug karo. $F = (8.99 \times 1 0^{9}) \frac{( 5 \times 1 0 ^{- 6} ) ( 2 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}} .$

Step 3 — arithmetic. Fraction ka numerator: $(5 \times 1 0^{- 6}) (2 \times 1 0^{- 6}) = 10 \times 1 0^{- 12} = 1.0 \times 1 0^{- 11}$ . Denominator: $(0.20)^{2} = 0.04$ . $F = 8.99 \times 1 0^{9} \times \frac{1.0 \times 1 0 ^{- 11}}{0.04} = 8.99 \times 1 0^{9} \times 2.5 \times 1 0^{- 10} = 2.25 N .$

Answer: $F \approx 2.25 N$ , repulsive (dono positive hain).

L2.2

Hydrogen atom mein ek electron aur ek proton approximately $r = 5.3 \times 1 0^{- 11} m$ door hain (Bohr radius). Dono ka charge magnitude $e = 1.6 \times 1 0^{- 19} C$ hai. Unke beech electric force nikalo.

Recall Solution

Step 1. Magnitudes: $∣ q_{1} ∣ = ∣ q_{2} ∣ = 1.6 \times 1 0^{- 19} C$ .

Step 2. $F = (8.99 \times 1 0^{9}) \frac{( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 5.3 \times 1 0 ^{- 11} ) ^{2}} .$

Step 3. Numerator: $(1.6 \times 1 0^{- 19})^{2} = 2.56 \times 1 0^{- 38}$ . Denominator: $(5.3 \times 1 0^{- 11})^{2} = 2.809 \times 1 0^{- 21}$ . $F = 8.99 \times 1 0^{9} \times \frac{2.56 \times 1 0 ^{- 38}}{2.809 \times 1 0 ^{- 21}} \approx 8.2 \times 1 0^{- 8} N .$

Answer: $F \approx 8.2 \times 1 0^{- 8} N$ , attractive (opposite signs). Roz ki zindagi mein yeh bahut chhota lagta hai, lekin yeh atom ko jode rakhta hai.

Level 3 — Analysis

Inke liye, positions matter karti hain, isliye hum direction ke saath kaam karte hain, Superposition Principle use karke: ek charge par net force baaki sab charges ki individual Coulomb forces ka vector sum hai, jahan har ek aisa compute kiya jaata hai jaise baaki wahan hain hi nahi.

Neeche ki figure L3.1 setup karti hai. Teen equal positive charges ek horizontal line par baithe hain, evenly spaced; red arrows middle charge $B$ par forces hain — ek right pointing (from $A$ ) aur ek left pointing (from $C$ ). Dekho ki arrows same length ke hain aur opposite directions mein hain: yeh visual balance exactly wahi cancellation hai jo algebra confirm karega.

Figure — Coulomb's law — force, comparison with gravity

L3.1

Teen charges ek line par hain (upar ki figure dekho). $q_{A} = + 3 μ C$ at $x = 0$ , $q_{B} = + 3 μ C$ at $x = 0.10 m$ , aur $q_{C} = + 3 μ C$ at $x = 0.20 m$ . Middle charge $q_{B}$ par net force nikalo.

Recall Solution

Step 1 — $q_{B}$ par pushes list karo. Do forces act karti hain: $A$ se (left par) aur $C$ se (right par). Dono positive-positive hain → repulsive. Figure mein yeh do red arrows hain.

$A$ , $B$ ko repel karta hai → $B$ ko right ( $+ x$ ) ki taraf push karta hai.
$C$ , $B$ ko repel karta hai → $B$ ko left ( $- x$ ) ki taraf push karta hai.

Step 2 — sizes. Dono source charges $3 μ C$ hain aur $B$ se same distance $0.10 m$ par hain: $F_{A B} = F_{C B} = (8.99 \times 1 0^{9}) \frac{( 3 \times 1 0 ^{- 6} ) ^{2}}{( 0.10 ) ^{2}} = (8.99 \times 1 0^{9}) \frac{9 \times 1 0 ^{- 12}}{0.01} = 8.09 N .$

Step 3 — vector sum. Yeh size mein equal hain lekin opposite direction mein hain (do equal-length red arrows), isliye cancel ho jaate hain: $F_{net} = + 8.09 - 8.09 = 0 N .$

Answer: middle charge zero net force feel karta hai — yeh ek symmetry point par baitha hai. (Lekin yeh line ke saath unstable hai: ise thoda sa hilao aur ek force jeet jaayegi.)

L3.2

Ab L3.1 mein middle charge ko $q_{B} = - 3 μ C$ (negative) kar do, $A$ aur $C$ ko $+ 3 μ C$ par rakhte hue. Ab $B$ par net force kya hai?

Recall Solution

Step 1 — directions flip ho jaati hain. Ab $A$ aur $B$ unlike hain → attract; $B$ , $A$ ki taraf khinchta hai, yaani left ki taraf. Isi tarah $C$ , $B$ ko right ki taraf attract karta hai. (Figure mein, mentally dono red arrows ko reverse karo.) Step 2 — sizes unchanged. Magnitude $∣ q_{1} q_{2} ∣$ use karti hai, aur $∣ - 3 ∣ = 3$ , isliye har force abhi bhi $8.09 N$ hai. Step 3 — sum. Phir equal aur opposite: $F_{net} = - 8.09 + 8.09 = 0 N .$ Answer: abhi bhi zero. $B$ ka sign flip karne se dono forces ek saath reverse ho gayi, isliye unka cancellation bana rehta hai. Is baar equilibrium line ke saath stable hai (left ki taraf nudge karo toh rightward pull center ki taraf badh jaata hai).

L3.3

Do equal charges $q = + 4 μ C$ , $(0, 0)$ aur $(0.30 m, 0)$ par hain. Ek teesra charge $Q = + 1 μ C$ , $(0.15 m, 0.20 m)$ par hai, midpoint ke exactly upar. $Q$ par net force ki magnitude aur direction nikalo.

Recall Solution

Step 1 — geometry. $Q$ midpoint ke upar hai, isliye yeh dono lower charges se equal distance par hai. Har ek se horizontal offset $= 0.15 m$ , vertical offset $= 0.20 m$ . $r = 0.1 5^{2} + 0.2 0^{2} = 0.0225 + 0.04 = 0.0625 = 0.25 m .$

Step 2 — har force ki size. $F = (8.99 \times 1 0^{9}) \frac{( 4 \times 1 0 ^{- 6} ) ( 1 \times 1 0 ^{- 6} )}{( 0.25 ) ^{2}} = (8.99 \times 1 0^{9}) \frac{4 \times 1 0 ^{- 12}}{0.0625} = 0.575 N .$ Har lower charge $Q$ ko repel karta hai (sab positive hain), use us charge se $Q$ tak ki line ke saath upar aur baahir push karta hai.

Step 3 — components mein split karo. Kyun: do arrows jo alag-alag directions mein point kar rahe hain unhe add karne ke liye, har ek ko horizontal ( $x$ ) aur vertical ( $y$ ) pieces mein todo, aur like-with-like add karo. Geometry se direction cosines: horizontal fraction $= 0.15/0.25 = 0.6$ , vertical fraction $= 0.20/0.25 = 0.8$ .

Left charge $Q$ ko upar-aur-right push karta hai: $F_{x} = + 0.575 (0.6) = + 0.345$ , $F_{y} = + 0.575 (0.8) = + 0.460$ .
Right charge $Q$ ko upar-aur-left push karta hai: $F_{x} = - 0.345$ , $F_{y} = + 0.460$ .

Step 4 — add karo. $F_{x, net} = + 0.345 - 0.345 = 0, F_{y, net} = 0.460 + 0.460 = 0.920 N .$ Symmetry se horizontal parts cancel ho jaate hain aur vertical parts reinforce karte hain.

Answer: $F_{net} \approx 0.92 N$ , seedha upar ( $+ y$ ) ki taraf point karta hai, pair se door.

Level 4 — Synthesis

L4.1

Ek line par, $q_{1} = + 9 μ C$ , $x = 0$ par hai aur $q_{2} = + 4 μ C$ , $x = 0.50 m$ par hai. Line par kahan ek teesra charge rakha jaaye taaki woh zero net force feel kare?

Neeche ki figure woh answer dikhati hai jo hum dhundh rahe hain. Red dot balance ("null") point mark karta hai; wahan do chhote arrows hain jo $q_{1}$ (blue) aur $q_{2}$ (green) ke opposing pushes hain. Dekho ki null point chhote charge $q_{2}$ ke kareeb baitha hai — uski kam distance uski kam charge ki kami poori karti hai. Do grey brackets un do distances ko label karte hain jo balance equation satisfy karni chahiye.

Recall Solution

Step 1 — kahan balance ho sakta hai? Test charge $q_{3}$ ko position $x$ par rakh do. Do repulsions (ya do attractions — same reasoning) cancel hone ke liye, unhe opposite directions mein hona chahiye aur equal size ka hona chahiye.

Charges ke beech ( $0 < x < 0.50$ ): $q_{1}$ right push karta hai, $q_{2}$ left push karta hai — yeh oppose karte hain, isliye cancellation yahan possible hai (yeh woh region hai jo figure mein dikhaya gaya hai).
Baahir ( $q_{1}$ ke left ya $q_{2}$ ke right): dono same direction mein push karte hain, cancel nahi ho sakte.

Toh unke beech dhundho.

Step 2 — magnitudes equal set karo. $q_{1}$ se distance $x$ hone do, toh $q_{2}$ se distance $(0.50 - x)$ hai. Unknown $q_{3}$ aur $k$ dono sides par appear karte hain aur cancel ho jaate hain: $k \frac{q _{1} q _{3}}{x ^{2}} = k \frac{q _{2} q _{3}}{( 0.50 - x ) ^{2}} \Rightarrow \frac{9 \times 1 0 ^{- 6}}{x ^{2}} = \frac{4 \times 1 0 ^{- 6}}{( 0.50 - x ) ^{2}} .$

Step 3 — solve karo. Cross-multiply karo aur square root lo (dono distances positive hain, isliye $+$ root lo): $\frac{9}{x ^{2}} = \frac{4}{( 0.50 - x ) ^{2}} \Rightarrow \frac{3}{x} = \frac{2}{0.50 - x} \Rightarrow 3 (0.50 - x) = 2 x .$ $1.5 - 3 x = 2 x \Rightarrow 1.5 = 5 x \Rightarrow x = 0.30 m .$

Answer: $x = 0.30 m$ par — chhote charge ke kareeb (figure se match karta hai), isliye weaker source ki kam distance uske kam charge ki kami poori karti hai. Result $q_{3}$ ke sign aur size se independent hai.

Level 5 — Mastery

L5.1

Char equal charges $q = + 2 μ C$ ek square ke corners par hain jiska side $a = 0.10 m$ hai. Ek corner par charge par net force ki magnitude aur direction nikalo.

Figure corner $P$ (red) aur us par act karne wali teen forces dikhati hai: do edges ke saath (blue aur green, har ek size $F_{a}$ ka) aur ek diagonal ke saath (orange, weaker kyunki diagonal charge zyada door hai). Bhaari red arrow unka vector sum hai — diagonal ke saath outward point karta hai, square ke centre se door. Dekho ki diagonal force dono edge forces ko reinforce karta hai na ki kuch cancel karta hai, isliye net kisi bhi single push se bada hai.

Recall Solution

Step 1 — teen pushes identify karo. Corner $P$ ko bottom-left par lo. Baaki teen charges hain: bottom ke saath adjacent corner par (distance $a$ ), side ke upar adjacent corner par (distance $a$ ), aur diagonal corner par (distance $a 2$ ). Sab $P$ ko repel karte hain, use har source se door push karte hain — yaani overall down-left (figure mein teen coloured arrows).

Step 2 — do adjacent forces. $F_{a} = k \frac{q ^{2}}{a ^{2}} = (8.99 \times 1 0^{9}) \frac{( 2 \times 1 0 ^{- 6} ) ^{2}}{( 0.10 ) ^{2}} = (8.99 \times 1 0^{9}) \frac{4 \times 1 0 ^{- 12}}{0.01} = 3.596 N .$ Ek $P$ ko seedha left push karta hai (bottom-right neighbour se), doosra seedha down push karta hai (top-left neighbour se). $P$ ko origin par rakho aur square first quadrant mein; toh yeh do forces $- x$ aur $- y$ ki taraf point karti hain, har ek $3.596 N$ ki.

Step 3 — diagonal force. Distance $a 2 = 0.10 2 = 0.1414 m$ , toh $r^{2} = 2 a^{2} = 0.02$ : $F_{d} = k \frac{q ^{2}}{2 a ^{2}} = \frac{F _{a}}{2} = 1.798 N .$ Yeh far corner se $P$ ki taraf point karta hai, yaani diagonal ke saath third quadrant mein: components $(- \frac{1}{2}, - \frac{1}{2})$ apni size ke saath, $F_{d, x} = F_{d, y} = - 1.798/ 2 = - 1.271$ har ek deta hai.

Step 4 — components add karo. $F_{x} = - 3.596 - 1.271 = - 4.867 N, F_{y} = - 3.596 - 1.271 = - 4.867 N .$ Do adjacent forces axes ke saath hain; diagonal wala evenly split hota hai, dono ko reinforce karta hai — toh symmetry se $F_{x} = F_{y}$ , outward diagonal ke saath point karte hain.

Step 5 — magnitude. $F_{net} = F_{x}^{2} + F_{y}^{2} = 2 (4.867) = 6.88 N .$

Answer: $F_{net} \approx 6.88 N$ , diagonal ke saath outward directed hai, square ke centre se door (hamare setup mein $P$ ke below-left $4 5^{\circ}$ par).

Yaad rakhne wala compact form: $F_{net} = \frac{k q ^{2}}{a ^{2}} (2 + \frac{1}{2})$ . Check karo: $3.596 (1.9142) = 6.88$ . ✓

L5.2

Parent note ka ratio reuse karo: do electrons ke liye, $F_{e} / F_{g} \approx 4.2 \times 1 0^{42}$ , jahan $m_{e} = 9.11 \times 1 0^{- 31} kg$ . Ab do protons compare karo ( $m_{p} = 1.67 \times 1 0^{- 27} kg$ , same charge magnitude $e = 1.6 \times 1 0^{- 19} C$ ). Bina poori recomputation ke predict karo ki ratio roughly kaise change hoga, phir direct calculation se verify karo.

Recall Solution

Step 1 — kya same rehta hai, kya change hota hai. Electric force ka gravitational force se ratio hai $\frac{F _{e}}{F _{g}} = \frac{k e ^{2} / r ^{2}}{G m ^{2} / r ^{2}} = \frac{k e ^{2}}{G m ^{2}},$ jahan $G = 6.67 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}$ gravitational constant hai. Gaur karo ki distance $r$ cancel ho jaata hai (dono forces $1/ r^{2}$ shape share karte hain). Charge magnitude $e$ electron aur proton ke liye identical hai, aur $k, G$ constants hain. Sirf mass $m$ change hota hai. Toh ratio $1/ m^{2}$ ki tarah scale karta hai.

Step 2 — predict karo. Proton, electron se zyada bhaari hai: $\frac{m _{p}}{m _{e}} = \frac{1.67 \times 1 0 ^{- 27}}{9.11 \times 1 0 ^{- 31}} \approx 1833.$ Kyunki ratio $\propto 1/ m^{2}$ hai, electrons ko protons se replace karne par yeh $183 3^{2} \approx 3.36 \times 1 0^{6}$ factor se chhota ho jaata hai: $\frac{F _{e}}{F _{g}}_{proton} \approx \frac{4.2 \times 1 0 ^{42}}{3.36 \times 1 0 ^{6}} \approx 1.25 \times 1 0^{36} .$

Step 3 — directly verify karo. $\frac{k e ^{2}}{G m _{p}^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 6.67 \times 1 0 ^{- 11} ) ( 1.67 \times 1 0 ^{- 27} ) ^{2}} \approx 1.24 \times 1 0^{36} . ✓$ Quick prediction ( $1.25 \times 1 0^{36}$ ) aur full calculation ( $1.24 \times 1 0^{36}$ ) agree karte hain.

Answer: approximately $1.2 \times 1 0^{36}$ — abhi bhi astronomically large, lekin electrons ke liye roughly $1 0^{6}$ guna chhota, purely isliye kyunki protons bhaare hote hain. Distance kabhi nahi aayi, kyunki dono forces $1/ r^{2}$ shape share karte hain.

Recall Quick self-check ladder

Teen equal charges ke middle par zero net force? ::: Haan, symmetry se do equal opposite pushes cancel ho jaati hain. $+ 9 μ C$ aur $+ 4 μ C$ (0.50 m apart) ke beech null point? ::: $9 μ C$ charge se 0.30 m par (chhote wale ke kareeb). Equal charges ke square ke corner par net force? ::: $\frac{k q ^{2}}{a ^{2}} (2 + \frac{1}{2})$ , outward diagonal ke saath. $F_{e} / F_{g}$ ratio mein $r$ kyun appear nahi karta? ::: Dono forces $\propto 1/ r^{2}$ hain, isliye distance divide ho jaati hai. Proton-proton electric/gravity ratio? ::: Approximately $1.2 \times 1 0^{36}$ , yaani $(m_{p} / m_{e})^{2} \approx 1 0^{6}$ guna electron value se chhota.

Connections: Electric Field (force per unit charge — agli abstraction), Gauss's Law (jahan $4 π$ kaam aata hai), Newton's Law of Gravitation (identical $1/ r^{2}$ skeleton), Superposition Principle (jaise humne upar vectors sum kiye), Electric Potential Energy (is force ko distance par integrate karo).