Capacitance (farads) am I doing this right?

• phys-lexic
In summary, the conversation discusses creating a parallel plate capacitor using two flat plates of aluminum, a sheet of paper, a sheet of glass, and a slab of paraffin. The capacitance for each material is calculated using the formula (capacitance) = ((epsilon not)x(Area)x(k)) / (d). The values for the capacitance are 5.885E-7 farads for paper, 5.885E-8 farads for glass, and 3.363E-9 farads for paraffin. However, there was an error in the conversion of the area from centimeters squared to meters squared which resulted in incorrect values. The correct values are 5.885E-
phys-lexic
1. To make a parallel plate capacitor, you have available two flat plates of aluminum (area = 190 cm2), a sheet of paper (thickness = 0.10 mm, κ = 3.5), a sheet of glass (thickness = 2.0 mm, κ. = 7.0), and a slab of paraffin (thickness = 10.0 mm, κ = 2.0). Find each's capacitance; then enter the largest capacitance possible, of the three, in nF and the smallest in pF.

2. (capacitance) = ((epsilon not)x(Area)x(k)) / (d)

**d, or distance between the capacitors, would be the thickness of the paper/glass/paraffin right?**

-smallest = 3363 pF (paraffin)
-largest = 588.5 nF (paper)

when i enter the values on my webassign it says I am wrong... are the values correct and just converted wrong or am i totally off [i did converted all the given values to meters before plugging them into the equation]

phys-lexic said:

This is not what I get. I get different exponents. Can you show one example of how you calculated the capacitance?

in example, for paper i did

(Eo) (A) (k) (d)
(8.85E-12)x(1.9)x(3.5)... then divided by .0001 (thickness)

the equation i though to use was C = ( (Eo)x(A)x(k) ) / (d)

The error lies in the conversion of the area; 190 cm$$^{2}\neq$$ 1.9 m$$^{2}$$.

oh wow... i was in a rush and forgot it was area... hha thanks alot

1. What is capacitance?

Capacitance is a measure of an object's ability to store an electrical charge.

2. How is capacitance measured?

Capacitance is measured in units called farads (F), named after Michael Faraday.

3. What factors affect capacitance?

The factors that affect capacitance include the distance between the plates, the surface area of the plates, and the type of material between the plates.

4. How can I calculate capacitance?

Capacitance can be calculated using the formula C = Q/V, where C is capacitance, Q is charge, and V is voltage.

5. What are some real-life applications of capacitance?

Capacitance is used in many electronic devices, including radios, televisions, and computers. It is also used in power grids to help regulate voltage and in energy storage systems, such as batteries and capacitors.

• Introductory Physics Homework Help
Replies
2
Views
6K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
17
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
24K