In the question number 66, the charge on capacitors `C_(1)and C_(4)` are

Options

A. `4xx10^(-3)C, 12xx10^(-3)C`
B. `6xx10^(-3)C,12xx10^(-3)C`
C. `2xx10^(-3)C,4xx10^(-3)C`
D. `3xx10^(-3)C, 2xx10^(-3)C`

Explanation

To solve the problem regarding the charge on capacitors \( C_1 \) and \( C_4 \), we can follow these steps:

### Step 1: Understand the Configuration
We need to determine the charge on capacitors \( C_1 \) and \( C_4 \). From the previous question, we know that the charge on capacitors \( C_1, C_2, \) and \( C_3 \) is the same, denoted as \( Q \), while the charge on capacitor \( C_4 \) is denoted as \( Q' \).

### Step 2: Write the Equation for Total Charge
The total charge in the circuit can be expressed as:
\[ \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} = 600 \]
This implies that:
\[ Q \left( \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right) = 600 \]

### Step 3: Find the Equivalent Capacitance
The term \( \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \) represents the equivalent capacitance \( C_s \) of the series combination of \( C_1, C_2, \) and \( C_3 \). Therefore, we can write:
\[ Q = 600 \times C_s \]

### Step 4: Calculate the Charge \( Q \)
Assuming \( C_s \) is given as \( \frac{20}{3} \, \mu F \):
\[ Q = 600 \times \frac{20}{3} \, \mu C = 4000 \, \mu C = 4 \times 10^{-3} \, C \]

### Step 5: Find the Charge \( Q' \) on Capacitor \( C_4 \)
For capacitor \( C_4 \), we can express the charge as:
\[ \frac{Q'}{C_4} = 600 \]
This leads to:
\[ Q' = C_4 \times 600 \]

### Step 6: Substitute the Value of \( C_4 \)
Assuming \( C_4 = 200 \, \mu F \):
\[ Q' = 200 \times 600 \, \mu C = 120000 \, \mu C = 12 \times 10^{-3} \, C \]

### Final Answers
- Charge on capacitor \( C_1 \): \( Q = 4 \times 10^{-3} \, C \)
- Charge on capacitor \( C_4 \): \( Q' = 12 \times 10^{-3} \, C \)