basic one on redistribution of charges.

JEE Advanced Online › Electricity and magnetism questions › basic one on redistribution of charges.

In the fig the capacitors C₁ and C₂ are at first charged to a potential of Î”V but with opposite polarity,so that points a and c are on the side of the respective positive plates of C₁ and C₂ and points b and d are on the side of the respective negative plates.Switches S₁ and S₂ are now closed.

A)What is the potential between points e and f?

B)What is the charge on C₁ and C₂?

#Physics #Electricity and magnetism

UP 0 DOWN 0 1 3
Post on FacebookPost anonymously?

3 Answers

1

decoder ·2009-11-03 02:54:47

V=\frac{\Delta V(C_{1}-C_{2})}{C_{1}+C_{2}}

Q_{1}=C_{1}V

Q_{2}=C_{2}V

UP 0 DOWN 0
Post on FacebookPost anonymously?

3

msp ·2009-11-03 02:58:46

dude i dun have the ans,can u give the soln.

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

decoder ·2009-11-03 03:09:21

initial charges

Q₁=C₁Î”V
and Q₂=C₂Î”V

when the capacitors are connected they will acquire a common potential V

and charge will remain conserve so
q_total=q₁-q₂ (since they are joined with opposite polarity)

so V=q_total/C_eq.

V= q₁-q₂ / C₁+C₂

just put values

and finally Q'₁=C₁V

and Q'₂=C₂V

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω