Two condensers are in parallel and the energy of the combination is `0.1 J`, when the difference. of potential between terminals is `2 V`. With the same two condensers in series, the energy `1.6 xx 10^-2` J for the same difference of potential across the series combination. What are the capacities?

Explanation

To solve the problem step by step, we will first analyze the given information and then apply the relevant formulas for capacitors in parallel and series.

### Step 1: Understand the given data
- Energy in parallel combination, \( U_p = 0.1 \, J \)
- Energy in series combination, \( U_s = 1.6 \times 10^{-2} \, J \)
- Voltage across both combinations, \( V = 2 \, V \)

### Step 2: Write the formula for energy in terms of capacitance
The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]

### Step 3: Set up the equations for the two configurations
#### For the parallel combination:
The equivalent capacitance for capacitors in parallel is:
\[ C_{eq,p} = C_1 + C_2 \]
Using the energy formula:
\[ U_p = \frac{1}{2} C_{eq,p} V^2 \]
Substituting the known values:
\[ 0.1 = \frac{1}{2} (C_1 + C_2) (2^2) \]
This simplifies to:
\[ 0.1 = \frac{1}{2} (C_1 + C_2) \cdot 4 \]
\[ 0.1 = 2 (C_1 + C_2) \]
Dividing both sides by 2:
\[ C_1 + C_2 = 0.05 \quad \text{(Equation 1)} \]

#### For the series combination:
The equivalent capacitance for capacitors in series is:
\[ \frac{1}{C_{eq,s}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Using the energy formula:
\[ U_s = \frac{1}{2} C_{eq,s} V^2 \]
Substituting the known values:
\[ 1.6 \times 10^{-2} = \frac{1}{2} C_{eq,s} (2^2) \]
This simplifies to:
\[ 1.6 \times 10^{-2} = \frac{1}{2} C_{eq,s} \cdot 4 \]
\[ 1.6 \times 10^{-2} = 2 C_{eq,s} \]
Dividing both sides by 2:
\[ C_{eq,s} = 0.008 \quad \text{(Equation 2)} \]

### Step 4: Relate the two equations
From Equation 2, we know:
\[ C_{eq,s} = \frac{C_1 C_2}{C_1 + C_2} \]
Substituting \( C_1 + C_2 = 0.05 \) from Equation 1:
\[ 0.008 = \frac{C_1 C_2}{0.05} \]
Multiplying both sides by 0.05:
\[ 0.0004 = C_1 C_2 \quad \text{(Equation 3)} \]

### Step 5: Solve the equations
Now we have two equations:
1. \( C_1 + C_2 = 0.05 \)
2. \( C_1 C_2 = 0.0004 \)

Let \( C_1 = x \) and \( C_2 = 0.05 - x \). Substituting into Equation 3:
\[ x(0.05 - x) = 0.0004 \]
Expanding and rearranging:
\[ 0.05x - x^2 = 0.0004 \]
\[ x^2 - 0.05x + 0.0004 = 0 \]

### Step 6: Use the quadratic formula to find \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = -0.05 \), \( c = 0.0004 \)
\[ x = \frac{0.05 \pm \sqrt{(-0.05)^2 - 4 \cdot 1 \cdot 0.0004}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{0.05 \pm \sqrt{0.0025 - 0.0016}}{2} \]
\[ x = \frac{0.05 \pm \sqrt{0.0009}}{2} \]
\[ x = \frac{0.05 \pm 0.03}{2} \]
Calculating the two possible values:
1. \( x = \frac{0.08}{2} = 0.04 \)
2. \( x = \frac{0.02}{2} = 0.01 \)

### Step 7: Identify \( C_1 \) and \( C_2 \)
Thus, we have:
- \( C_1 = 0.04 \, F \)
- \( C_2 = 0.01 \, F \)

### Final Answer
The capacitances are:
- \( C_1 = 0.04 \, F \)
- \( C_2 = 0.01 \, F \)

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