Question

In the second part of the lab activities, students charged another capacitor (1) using a 165 V

source. Once charged, the capacitor was then removed from removed from the source and
then connected to another capacitor (2) which was initially uncharged. If the students
measured the final potential difference across each capacitor to be 15V, determine the value
of (2).

Answer 1

To determine the value of capacitor (2), we can use the principle of conservation of charge.

According to the principle of conservation of charge, the total charge before and after the capacitors are connected should be the same.

The charge stored in a capacitor can be calculated using the formula:

Q = CV

Where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.

Let's assume the capacitance of capacitor (1) is C1 and the capacitance of capacitor (2) is C2.

Initially, capacitor (1) is charged to 165V. Therefore, the charge stored in capacitor (1) is Q1 = C1 * 165.

When capacitor (1) is connected to capacitor (2), the charge is shared between them. Since the final potential difference across each capacitor is 15V, the total charge stored in both capacitors is the same.

Therefore, we have:

Q1 = Q2

C1 * 165 = C2 * 15

To find the value of C2, we can rearrange the equation:

C2 = (C1 * 165) / 15

The value of C1 is not provided in the given information, so we cannot calculate the exact value of C2 without that information.

Step-by-step explanation: