What is the maximum charge on plates when dielectric removed

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Homework Help Overview

The problem involves a capacitor with parallel plates and a dielectric material, specifically nylon, which breaks down under a certain voltage. The question focuses on determining the maximum charge the plates can hold when the dielectric is removed and replaced with air.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the maximum charge using the relationship between capacitance, voltage, and electric field strength. Some participants question the unit conversions and calculations related to area and capacitance.

Discussion Status

The discussion is ongoing, with participants providing guidance on unit conversions and identifying potential errors in the calculations. There is a recognition of a misunderstanding regarding the area conversion from square centimeters to square meters.

Contextual Notes

There is a specific focus on ensuring correct unit conversions, particularly for area and electric field strength, which are critical to the calculations involved in the problem.

akirez
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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?
 
Last edited:
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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?
 
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.
 
What did you get for the area after you converted to m2?
 
88.5 cm^2 = 0.885 m^2
 
akirez said:
88.5 cm^2 = 0.885 m^2
This is not correct. Note that you are converting square centimeters.
 
Wow. That flew right over my head. That was my problem. Thanks so much for your help!
 
Good work.
 

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