Capacitor C₁ is connected alone to a battery and charged until the magnitude of the charge on each plate is 4.0 * 10^-⁸C.Then it is removed from the battery and connected to two other capacitors C₂and C₃, as shown.The charge on the positive plate of C₁ is then 1.0 *10^-⁸C.The charges on the positive plates of C₂ and C₃are: A)q₂ = 3.0*10 ^-8C and q₃ = 3.0 *10 ^- ⁸C B)q₂ = 2.0 *10 ^- ⁸ C and q₃ = 2.0 *10 ^- ⁸C C)q₂ = 5.0 * 10^-⁸ C and q₃ = 1.0 * 10^-⁸C D)q₂ = 3.0* 10^-⁸C and q₃ = 1.0 * 10^-⁸C E)q₂ = 1.0 *10^-⁸ C and q₃ = 3.0 *10^-⁸C

Question

Quizplus · Accepted Answer

We can use the conservation of charge in a series combination of capacitors. Since the capacitors are connected in series, the magnitude of the charge on each capacitor will be the same. Let q be the magnitude of the charge on each capacitor. Before the capacitors are connected, the charge on each plate of C₁ is 4.0 F1F1F1S1F1F1F10 10F1F1F1S1^F1F1F10⁸C. Therefore, the total charge on C₁ is 8q. After the capacitors are connected in series, the charge on the positive plate of C₁ is 1.0 F1F1F1S1F1F1F10 10F1F1F1S1^F1F1F10⁸C. Therefore, the total charge on C₁ is 2q. By conservation of charge, the total charge on C₂ and C₃ combined is 6q. Since the capacitors are in series, the charge on the positive plate of C₂ is q/3 and the charge on the positive plate of C₃ is 5q/3 (or vice versa). Therefore, the answer is (A) q₂ = 3.0 F1F1F1S1F1F1F10 10F1F1F1S1^F1F1F10^-8C and q₃ = 3.0 F1F1F1S1F1F1F10 10F1F1F1S1^F1F1F10⁸C.

Capacitor C1 Is Connected Alone to a Battery and Charged

Capacitor C₁ Is Connected Alone to a Battery and Charged