A 100pF capacitor is charged to a potential difference of 80.0V and the charging battery is disconnected. The capacitor is then connected in parallel with a second capacitor. If the potential difference across the first capacitor drops to 35.0 V what is the capacitance of this second capacitor?

Explanation

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

### Step 1: Calculate the initial charge on the first capacitor (C1)
The first capacitor (C1) has a capacitance of 100 pF (picoFarads) and is charged to a potential difference of 80 V. The charge (Q) on a capacitor is given by the formula:

\[ Q = C \times V \]

Where:
...