Is My Calculation of the Radius in a Dielectric Strength Problem Correct?

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SUMMARY

The discussion focuses on calculating the radius of a parallel plate capacitor filled with a dielectric material, specifically one with a dielectric constant of 3.5 and a dielectric strength of 1.4 x 107 N/C. The user correctly applies the equation E = Q/(ε0kA) to find the area A, using the charge Q of 1.7 x 10-7 C. The correct formula for the radius r is derived from A = πr2, leading to a final radius of 2.8 x 10-4 m.

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  • Knowledge of the constants ε0 (vacuum permittivity) and dielectric constant
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Electrical engineers, physics students, and anyone involved in the design and analysis of capacitors and dielectric materials will benefit from this discussion.

kris24tf
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Dielectric Strength problem. Please Help! :(

If anyone can let me know if I went about this right, I'd appreciate it, as well as any advice.

The dielectric strength of an insulating material is the maximum electric field strength to which the material can be subjected without electrical breakdown occurring. Suppose a parallel plate capacitor is filled with a material whose dielectric constant is 3.5 and whose dielectric strength is 1.4 E 7 N/C. If this capacitor is to store 1.7 E -7 C of charge on each plate without suffering breakdown, what must be the radius of its circular plates?

I started with the equation E=Q/Eo*k*A, then found that A=Q/Eo*k*E.

I then plugged the numbers into this equation to get 2.3E-7/8.85E-12(3.5)(1.4E7).

Is this the right way to do this problem or do I need to do something differently, as I'm not sure if the radius they're looking for could be found with the Area I got?
 
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The answer should be 2.8E-4 m.To get the radius, you need to use A = πr2. So, rearrange the equation to solve for r and you will get:r = √(A/π)Plug in your value for A and you will get r = 2.8E-4 m.
 

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