Six `1 mu F` capacitors are so arranged that their equivalent capacitance is `0.70 mu F`. If a potential difference of 600 volt is applied to the combination, what charge will appear on each capacitor ?

Answer

To solve the problem step by step, we will analyze the arrangement of the capacitors and apply the relevant formulas for capacitors in series and parallel.

### Step 1: Understanding the Arrangement
We have six capacitors, each with a capacitance of \(1 \mu F\). The equivalent capacitance of the arrangement is given as \(0.7 \mu F\). We need to determine how these capacitors are arranged to achieve this equivalent capacitance.

### Step 2: Finding the Charge on the Combination
Using the formula for charge, we can find the total charge \(Q\) on the combination when a potential difference \(V\) is applied:

\[ Q = C_{eq} \times V \]

Where:
- \(C_{eq} = 0.7 \mu F\)
- \(V = 600 V\)

Calculating the charge:

\[ Q = 0.7 \times 10^{-6} F \times 600 V = 420 \times 10^{-6} C = 420 \mu C \]

### Step 3: Analyzing the Series and Parallel Configuration
Assuming a configuration where three capacitors are in series and two capacitors are in parallel with this series combination, we can analyze the voltage across each section.

### Step 4: Finding Voltage Across Series Capacitors
In a series configuration, the charge is the same across all capacitors. Therefore, the charge on the three capacitors in series is \(Q = 420 \mu C\).

To find the voltage across these three capacitors, we use:

\[ V_{series} = \frac{Q}{C_{series}} \]

Where \(C_{series}\) for three capacitors in series is:

\[ \frac{1}{C_{series}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \implies C_{series} = \frac{C}{3} = \frac{1 \mu F}{3} = \frac{1}{3} \mu F \]

Calculating the voltage across the series capacitors:

\[ V_{series} = \frac{420 \mu C}{\frac{1}{3} \mu F} = 420 \times 3 = 1260 V \]

### Step 5: Finding Voltage Across Parallel Capacitors
Since the total voltage is \(600 V\), the voltage across the three capacitors in series is \(V_{series} = 420 V\). The remaining voltage across the two parallel capacitors is:

\[ V_{parallel} = 600 V - 420 V = 180 V \]

### Step 6: Finding Charge on Each Parallel Capacitor
The charge on each of the two capacitors in parallel is given by:

\[ Q_{parallel} = C \times V_{parallel} = 1 \mu F \times 180 V = 180 \mu C \]

### Step 7: Finding Charge on Each Series Capacitor
The charge on the three capacitors in series is the same:

\[ Q_{series} = 420 \mu C \]

### Final Summary of Charges
- Charge on each of the three capacitors in series: \(420 \mu C\)
- Charge on each of the two capacitors in parallel: \(180 \mu C\)

### Conclusion
The charges on the capacitors are:
- Three capacitors in series: \(420 \mu C\) each
- Two capacitors in parallel: \(180 \mu C\) each