Dielectric Help: Solving Parallel Plate Capacitor

[Bob Loblaw]

[SOLVED]Dielectric help

Homework Statement

To make a parallel plate capacitor, you have available two flat plates of aluminum (area = 180 cm2), a sheet of paper (thickness = 0.10 mm, k= 3.5), a sheet of glass (thickness = 2.0 mm, . k= 7.0), and a slab of paraffin (thickness = 10.0 mm, k= 2.0).
(a) What is the largest capacitance possible using one of these dielectrics?
_________nF
(b) What is the smallest?
________pF

Homework Equations

$$C=k\epsilon_0 \frac{A}{d}$$

The Attempt at a Solution

I used the above equation to evaluate the capacitance of each in this manner:
1) 3.5*8.85e-12*.180m^2/.0001 = 5.5755e-8
2) 7.0*8.85e-12*.180m^2/.002 = 5.5755e-9
3) 2.0*8.85e-12*.180m^2/.010 = 3.186e-10

I recorded the largest as .55.755 and the smallest as 318.6 (in their appropriate units) both of which were wrong. Any ideas where I messed up?

 

[Bob Loblaw]

self-solved it. 180cm2 is .018m2, not .180m2 as I have recorded.

 

[Moetasim]

Thank you for sharing your solution attempt. It seems like you have correctly applied the formula for capacitance using the given dielectrics and dimensions. However, it is important to note that the units for capacitance are Farads, not nF or pF. Therefore, your answers should be in the form of 55.755 nF and 318.6 pF respectively. Also, make sure to double check your calculations to ensure accuracy. Keep up the good work!