Answer:
Charge on C₁ = charge on all the three capacitors in series with it = 7.5 μC
Explanation:
Since the same voltage in the battery is used for the entire rundown,
From this information "only C₁ is connected to the battery, the charge on C₁ is 30.0 μC",
Q = C₁V = 30 μC
V = (30/C₁)
the series combination of C₂ and C₁ is connected across the battery, the charge on C₁ is 15.0 μC
The charge on both capacitors are the same and equal to 15 μC (because they are in series)
Q = (Ceq) V = 15 μC
(Ceq) = (15/V) μF
The voltage is still the same as in the first connection process
V = (30/C₁)
(Ceq) = (15/V) μF
(Ceq) = 15 ÷ (30/C₁)
(Ceq) = 15 × (C₁/30) = 0.5 C₁
(1/Ceq) = (2/C₁)
For series connection
(1/Ceq) = (1/C₁) + (1/C₂)
(2/C₁) = (1/C₁) + (1/C₂)
(2/C₁) - (1/C₁) = (1/C₂)
(1/C₁) = (1/C₂)
C₁ = C₂
C₂ = C₁
C₃, C₁, and the battery are connected in series, resulting in a charge on C₁ of 10.0 μC.
The charge on both capacitors are the same and equal to 10 μC (because they are in series)
Q = (Ceq) V = 10 μC
(Ceq) = (10/V) μF
The voltage is still the same as in the first connection process
V = (30/C₁)
(Ceq) = (10/V) μF
(Ceq) = 10 ÷ (30/C₁)
(Ceq) = 10 × (C₁/30) = 0.333 C₁
(1/Ceq) = (3/C₁)
For series connection
(1/Ceq) = (1/C₁) + (1/C₃)
(3/C₁) = (1/C₁) + (1/C₃)
(3/C₁) - (1/C₁) = (1/C₃)
(2/C₁) = (1/C₃)
C₁ = 2C₃
C₃ = (C₁/2)
C₁, C₂, and C₃ are connected in series with one another and
with the battery, what is the charge on C₁
The charge on C₁ is the same as the charge on all the capacitors and equal to Q,
Q = (Ceq) V
(1/Ceq) = (1/C₁) + (1/C₂) + (1/C₃)
Substituting for C₂ and C₃
C₂ = C₁ and C₃ = (C₁/2)
(1/C₂) = (1/C₁) and (1/C₃) = (2/C₁)
(1/Ceq) = (1/C₁) + (1/C₁) + (2/C₁)
(1/Ceq) = (4/C₁)
Ceq = (C₁/4)
Q = (Ceq) V = (C₁/4) V
But recall that V = (30/C₁) from the first connection
Q = (C₁/4) (30/C₁)
Q = (30/4) = 7.5 μC
Hope this helps!
