What is the maximum charge on plates when dielectric removed

[akirez]

Homework Statement

A capacitor with two parallel plates of area 88.5 cm2 and nylon (dielectric strength 14 MV/m) inserted between them breaks down when a potential difference of 5.80 kV is applied. What is the maximum charge the plates will hold when the nylon is removed and the space filled with air (dielectric strength 3 MV/m)?

Homework Equations

Q = CΔV
ΔV = Ed
Co = εoA/d
Coκ = εoAκ/d
E = Eo/κ

The Attempt at a Solution

The maximum electric field with the nylon inserted is 14 MV/m. ΔV = Ed and we have 5800 V applied.
Solving for d gives d = 0.00041429 m.

The maximum electric field once the nylon removed is 3 MV/m. ΔV = Ed.
Solving for ΔV gives ΔV = 1242.87 V.

C = εoA/d gives C= 1.89*10^-8 F.

Q = CΔV = (1.89*10^-8 F)(1242.87 V) = 0.0000235 C.

The online assignment page is rejecting my answer. Where am I going wrong?

 

[TSny]

Welcome to PF!

Did you correctly convert cm² to m²?

Why did 10^-8 change to 10^-12 for C in the calculation of Q?

 

[akirez]

Sorry the 10^-12 was a typo. Indeed when I did the calculation I used 10^-8 for C. Also I converted cm^2 to m^2 and converted MV and kV to volts.

 

[TSny]

What did you get for the area after you converted to m²?

 

[akirez]

88.5 cm^2 = 0.885 m^2

 

[TSny]

88.5 cm^2 = 0.885 m^2

This is not correct. Note that you are converting square centimeters.

 

[akirez]

Wow. That flew right over my head. That was my problem. Thanks so much for your help!

 

[TSny]

Good work.