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

In summary, the conversation discusses the concept of dielectric strength and its application in a parallel plate capacitor problem. The goal is to find the radius of the circular plates that can store 1.7 E -7 C of charge without experiencing breakdown. The expert suggests using the equation A = πr2 to solve for the radius, which results in a value of 2.8E-4 m.
  • #1
kris24tf
35
0
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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  • #2
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.
 
  • #3


Your approach to the problem is correct. The equation E=Q/Eo*k*A is the correct equation for calculating the electric field strength in a parallel plate capacitor. And finding the area using A=Q/Eo*k*E is also correct.

However, in order to find the radius of the circular plates, you will need to use the equation for the area of a circle, which is A=πr^2. This means that you will need to rearrange the equation to solve for the radius (r), which would be r=sqrt(A/π).

Once you have the value for the area, you can plug it back into the equation A=Q/Eo*k*E to find the radius. So your final equation would be r=sqrt((1.7E-7)/(8.85E-12*3.5*1.4E7*π)).

I would also recommend double checking your units to ensure they are consistent and to use scientific notation for easier calculations.

Additionally, it is always a good idea to double check your answer and do a quick sanity check to make sure it makes sense. In this case, the radius of the plates should be in meters, so if you get a value in a different unit, it may be a sign that you made a mistake in your calculations.

Overall, your approach and equations are correct, but just be sure to use the correct equation for finding the radius and to double check your units and answer.
 

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

1. What is dielectric strength?

Dielectric strength is a measure of the ability of a material to withstand an electric field without breaking down and becoming conductive.

2. What factors affect dielectric strength?

The main factors that affect dielectric strength are the material's composition, thickness, and temperature. Other factors that can influence it include the presence of impurities, humidity, and the rate of change of the electric field.

3. Why is dielectric strength important?

Dielectric strength is important because it is a key property in determining the safety and reliability of electrical systems. It is also crucial in the design and testing of insulating materials for various applications, such as in capacitors and cables.

4. How is dielectric strength measured?

Dielectric strength is typically measured by applying an increasing electric field to a sample of the material until it breaks down and becomes conductive. The maximum electric field that the material can withstand before breakdown occurs is then recorded as its dielectric strength.

5. What are some common materials with high dielectric strength?

Some common materials with high dielectric strength include air, glass, porcelain, and certain types of plastic. Additionally, materials such as mica, rubber, and ceramic have high dielectric strength and are often used in electrical insulation applications.

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