KYA: Hum pooch rahe hain ki potentials kaise combine hote hain. Kyun matter karta hai:V ek scalar hai, jo ek dot-product se bana hai jo pehle hi ek single number mein collapse ho chuka hai.
Answer: Tum seedhe numbers add karte ho (unke +/− signs ke saath), kabhi arrows nahi. Potential ka superposition hai Vtotal=∑irikqi — ordinary algebraic addition.
Direction sirf E=−∇V mein hoti hai, V mein nahi.
Recall Solution L1·Q2
V/m kyun legal hai:E=ΔV/d se (uniform field), field = (volts) ÷ (metres) = V/m.
Aur 1V=1J/C=1N⋅m/C, isliye
1mV=1mN⋅m/C=1CN.
V/m aur N/C same unit hain. Dono correct hain.
V=rkq use karo (yeh ∞ se radially andar walk karne se derive hota hai jahan E∥dl):
V=0.20(8.99×109)(5×10−9)=224.75V.Sign check: positive charge ⇒ V>0, aur r ghatne par badhta hai. Correct.
Recall Solution L2·Q2
Uniform field ⇒ E integral se bahar aa jaata hai, isliye E=dΔV (dekho Capacitance and parallel-plate fields (E = ΔV/d)):
E=4.0×10−312=3000V/m.Direction:Ehigh-potential plate se low-potential plate ki taraf point karta hai
(V mein field "neeche ki taraf" jaati hai).
Recall Solution L2·Q3
E=−∇V — har direction mein scalar ko differentiate karo aur sign flip karo
(dekho Gradient operator ∇ and directional derivatives):
Ex=−∂xV=−8x,Ey=−∂yV=3z,Ez=−∂zV=3y.(1,2,1) par: E=(−8,3,6) V/m. Additive constant +7 vanish ho jaata hai — sirf V mein changes field produce karte hain.
Field dwara kaam =q(VA−VB) (from Work–energy theorem in electrostatics):
Wfield=(−3×10−6)(40−90)=(−3×10−6)(−50)=+1.5×10−4J.Analysis: number positive aaya — field ne motion mein help ki. Kyun? Ek negative charge zyada potential ki taraf push hoti hai (positive charge ke opposite), isliye bade V tak jaana uske liye "neeche" jaana hai. q ke sign ne sab kuch decide kiya.
Recall Solution L3·Q2
Scalar contributions add karo (superposition, koi arrows nahi). Dono charges M se 0.15 m dur hain:
= \frac{8.99\times10^{9}}{0.15}\left(4\times10^{-9} - 2\times10^{-9}\right). $$
$$ V_M = (5.993\times10^{10})(2\times10^{-9}) = 119.87\ \text{V}. $$
**Analysis:** negative charge *subtract* karta hai, isliye $V_M$ akele $+4$ nC se kam hai.
Yahi wajah hai ki $V$ zero ya negative ho sakta hai jahan bhi $\mathbf{E}\neq 0$ ho.Recall Solution L3·Q3
Maano point q1 se x distance par hai, to q2 se 0.30−x door hai. V=0 set karo:
\;\Rightarrow\; \frac{4}{x} = \frac{2}{0.30-x}. $$
Cross-multiply karo: $4(0.30 - x) = 2x \Rightarrow 1.20 = 6x \Rightarrow x = 0.20$ m.
**Analysis:** yahan $V=0$ hai, lekin $\mathbf{E}\neq 0$ — potential ek *sum* hai jo cancel ho sakta hai jabki vector field nahi hota. Zero potential ≠ zero field.
(a) Bahar (r>R): field point-like hai, to ∞ se integrate karne par milta hai
V(r)=−∫∞rr′2kQdr′=rkQ.r=R par surface value hai V(R)=RkQ.
(b) Andar (r<R): yahan E=0 hai, isliye aur andar jaane par integral mein kuch nahi judta:
V(r)=V(R)−∫Rr=0Edr′=V(R)=RkQ(constant).Kyun constant: zero field ka matlab zero slope hai, isliye V andar flat hai — poora interior ek equipotential hai.
Centre par value:V(0)=RkQ=0.10(8.99×109)(8×10−9)=719.2V.
Recall Solution L4·Q2
Saari potential energy kinetic energy mein convert ho jaati hai (from Work–energy theorem in electrostatics):
ΔKE=q(Vstart−Vend)=(1.6×10−19)(719.2−0).ΔKE=1.15072×10−16J.Sign check: positive charge, positive V, lower V ki taraf jaata hai — energy release hoti hai, KE positive. Sahi hai.
= -\frac{\lambda}{2\pi\varepsilon_0}\ln\!\frac{r_b}{r_a}. $$
**Infinity kyun fail hoti hai:** agar $r_a$ ya $r_b \to \infty$, to $\ln(\infty)$ **diverge** karta hai — integral ki koi finite value nahi hai. Isliye tum yahan $V=0$ at $\infty$ nahi rakh sakte; ek finite reference radius choose karo.
Sirf **difference** (ratio ka logarithm) physical hai. Yeh exactly parent ki infinite distributions ke baare mein warning hai.Recall Solution L5·Q2
Er=−drd(rkq)=−(−r2kq)=r2kq.
Yeh exactly Coulomb's field hai. Yeh kyun matter karta hai: do master relations V=−∫E⋅dl aur E=−∇V
perfect inverses hain — field integrate karo to V milta hai, V differentiate karo to field wapas milti hai. Minus signs cancel hokar q>0 ke liye correct outward-pointing field dete hain.
Recall Solution L5·Q3
Jhooth hai. Force E=−∇V (yaani slope) se aati hai, V ki value se nahi. x=0.20 m par:
dono charges ke fields same direction mein point karte hain us segment par (+q1 se dur aur −q2 ki taraf dono +x direction mein hain), isliye woh add hote hain, cancel nahi:
E=0.202k(4×10−9)+0.102k(2×10−9)=0.V=0 ek value hai; wahan V ka gradient bada hai. Zero potential, strong field — is poore page ka mastery lesson yahi hai.
