Question

1. In the first part of this lab, you will build your own parallel plate capacitor by with two aluminum foil sheets and on or more transparency sheets in between them.

1. If the each aluminum sheet is 10 cm x 10 cm, and a piece of transparency is ~0.1mm thick, what is the capacitance of the 2 foil sheets separated by 1 transparency sheet? Note: pay close attention to the units!

2. If the distance between two sheets is 0.1 mm, what area would they have to be so that the capacitance is 1 F?

3. If the area of each sheet is 10 cm x 10 cm, how close would they have to be so that the capacitance is 1 F? For reference, the diameter of a hydrogen atom is ~10^-10 m.

2. In this lab, you will be given 4 capacitors. Below are two blown-up schematics of the breadboard. Indicate how you will connect the 4 capacitors in series (left) and in parallel (right). Make sure your circuit is a closed loop; the voltage source has been already connected to the top rows. Remember: each hole can only connect firmly to a single wire.

For Parallel Connections $3_breadboard_schematic$

For Series Connections $3_breadboard_schematic$

Answer 1

Capacitance can be calculated by formula C = A epsilon_0/d where, C = Capacitance of capacitor A = Area of the sheets (plates) d = distance between sheets (plates) epsilon_0 = Electric permivity in space or vaccumm. (a) Area, A = 10 cm times 10 cm = 100 cm^2 A = 100 times 10^-4 m^2 = 10^-2 m^2 distance, d = 0.1 mm = 0.1 times 10^-3 m d = 10^-4 m C = A epsilon_0/d C = 10^-2 times 8.85 times 10^-12/10^-4 C = 10^-2 times 8.85 times 10^-12 times 10^+ 4 F C = 8.85 times 10^-10 F (b) Now, C = 1 F d = 0.1 mm = 10^-4 m C = A epsilon_0/d 1 = A times 8.85 times 10^-12/10^-4 A = 10^-4/8.85 times 10^-12 = 10^8/8.85 = 0.113 times 10^8 m^2 A = 11.3 times 10^6 m^2 \r\n (c) A = 10 times 10 cm = 10^-2 m^2 C = 1 F C =