Three parallel plate capacitors (C1,C2 and C3 ) are in series with a battery of 100 V. C1=2500 micro Farad; C2=2C1 , C3=3C1 .
(a) What is the potential difference over each capacitor and how much charge is on each capacitor?
Across C1 :V1 (in Volts) =
incorrect
Q1 (in C) =
incorrect
Across C2 :V2 (in Volts) =
incorrect
Q2 (in C) =
incorrect
Across C3 : V3 (in Volts) =
incorrect
Q3 (in C) =
incorrect
(b) We now connect a 4th parallel plate capacitor ( C4=4C1 ) to the battery. One side of the capacitor is connected to the negative side of the battery, the other side is connected to the positive side of the battery.
What is the potential difference over each of the 4 capacitors and what is the charge on each?
Across C1 : V1 (in Volts) =
incorrect
Q1 (in C) =
incorrect
Across C2 : V2 (in Volts) =
incorrect
Q2 (in C) =
incorrect
Across C3 : V3 (in Volts) =
incorrect
Q3 (in C) =
incorrect
Across C4 : V4 (in Volts) =
incorrect
Q4 (in C) =
incorrect
(c) We make one more change: We still have our 4 capacitors as above, but we now place a dielectric (K = 3) between the plates of the 4th capacitor.
What is the potential difference across C4 and what is the charge on it?
Across C4 : V4 (in Volts) =
incorrect
Q4 (in C) =
Please some one answer.
Dude, Edx is for you to solve on own. Not copy-pasting in forums for others to solve for you.
To solve this problem, we need to apply the principles of capacitors in series and parallel connections.
(a) Capacitors in series have the same charge (Q) on each capacitor, but the potential difference (V) is divided between them. The total capacitance (C_total) of capacitors in series is the reciprocal of the sum of their reciprocals.
In this case, C1, C2, and C3 are in series. Using the formula:
1/C_total = 1/C1 + 1/C2 + 1/C3
we can calculate C_total:
1/C_total = 1/2500μF + 1/5000μF + 1/7500μF
= 1/2500μF + 1/2500μF + 1/2500μF
= 3/2500μF
= 1/833.33μF
Therefore, C_total = 833.33μF.
Now, using the potential difference equation for capacitors in series:
V_total = V1 + V2 + V3
we can find V1, V2, and V3. Since the capacitors are in series, V_total is equal to the battery voltage, which is 100V.
V1 = V_total * (C1 / C_total)
= 100V * (2500μF / 833.33μF)
≈ 300V
V2 = V_total * (C2 / C_total)
= 100V * (2 * C1 / 833.33μF)
= 100V * (2 * (2500μF) / 833.33μF)
= 100V * (5000μF / 833.33μF)
≈ 600V
V3 = V_total * (C3 / C_total)
= 100V * (3 * C1 / 833.33μF)
= 100V * (3 * (2500μF) / 833.33μF)
= 100V * (7500μF / 833.33μF)
≈ 900V
For the charge on each capacitor:
Q1 = Q2 = Q3 = Q_total (since capacitors in series share the same charge)
Q_total = V_total * C_total
= 100V * 833.33μF
= 83333.3μC
Therefore, Q1 = Q2 = Q3 = 83333.3μC.
(b) When the 4th capacitor (C4 = 4C1) is connected in parallel to the battery, the potential difference across each capacitor remains the same as in part (a), but the total charge (Q_total) is divided between them.
The potential difference across each capacitor is still V1 ≈ 300V, V2 ≈ 600V, V3 ≈ 900V, and V4 ≈ 100V (since they are all connected to the battery voltage).
The total charge (Q_total) is still 83333.3μC.
The charge on each capacitor can be found using the equation for charge on capacitors in parallel:
Q_total = Q1 + Q2 + Q3 + Q4
Since C4 = 4C1, the charge on C4 will be four times that of C1:
Q4 = 4 * Q1
= 4 * 83333.3μC
≈ 333333.2μC
The charges on C1, C2, and C3 remain the same as in part (a), so Q1 = Q2 = Q3 = 83333.3μC.
(c) When a dielectric with a dielectric constant (K = 3) is placed between the plates of C4, the capacitance of C4 increases by a factor of K.
The potential difference across C4 remains the same as in part (b), V4 ≈ 100V.
The charge on C4 can be found using the capacitance equation:
C4_new = K * C4
= 3 * 4 * C1
= 12 * C1
Q4_new = C4_new * V4
= (12 * C1) * 100V
= 1200Q
Therefore, Q4_new ≈ 1200Q.