`A` 800 `pF` capacitor is charged by a `100 V` battery. After sometime, the battery is disconnected. The capacitor is then connected to another `800 pF` capacitor. What is the electrostatic energy stored ?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the charge stored in the first capacitor (C1)
The charge (Q1) stored in a capacitor is given by the formula:
\[ Q = C \times V \]
Where:
- \( C \) is the capacitance in farads
- \( V \) is the voltage in volts

Given:
- \( C1 = 800 \, \text{pF} = 800 \times 10^{-12} \, \text{F} \)
- \( V1 = 100 \, \text{V} \)

Calculating \( Q1 \):
\[ Q1 = C1 \times V1 = (800 \times 10^{-12} \, \text{F}) \times (100 \, \text{V}) = 8 \times 10^{-7} \, \text{C} \]

### Step 2: Understand the connection of the two capacitors
After disconnecting the battery, the charged capacitor (C1) is connected to an uncharged capacitor (C2). Since C2 is uncharged, its initial charge (Q2) is 0.

### Step 3: Calculate the equivalent capacitance when C1 and C2 are connected in parallel
When two capacitors are connected in parallel, their equivalent capacitance (C_eq) is given by:
\[ C_{eq} = C1 + C2 \]

Given:
- \( C2 = 800 \, \text{pF} = 800 \times 10^{-12} \, \text{F} \)

Calculating \( C_{eq} \):
\[ C_{eq} = C1 + C2 = (800 \times 10^{-12} \, \text{F}) + (800 \times 10^{-12} \, \text{F}) = 1600 \times 10^{-12} \, \text{F} = 1.6 \, \text{nF} \]

### Step 4: Calculate the common voltage (V) across the capacitors after connection
The total charge (Q) remains the same when the capacitors are connected:
\[ Q = Q1 + Q2 = Q1 + 0 = Q1 \]

The voltage across the equivalent capacitance is given by:
\[ Q = C_{eq} \times V \]
Thus,
\[ V = \frac{Q}{C_{eq}} \]

Substituting the values:
\[ V = \frac{8 \times 10^{-7} \, \text{C}}{1.6 \times 10^{-9} \, \text{F}} = 50 \, \text{V} \]

### Step 5: Calculate the electrostatic energy stored in the system
The energy (U) stored in a capacitor is given by:
\[ U = \frac{1}{2} C \times V^2 \]

Using the equivalent capacitance and the common voltage:
\[ U = \frac{1}{2} C_{eq} \times V^2 \]
\[ U = \frac{1}{2} (1.6 \times 10^{-9} \, \text{F}) \times (50 \, \text{V})^2 \]

Calculating:
\[ U = \frac{1}{2} \times 1.6 \times 10^{-9} \times 2500 \]
\[ U = 2 \times 10^{-6} \, \text{J} \]

### Final Answer
The electrostatic energy stored in the system is:
\[ U = 2 \, \mu J \]

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