How Do You Calculate Parameters of a Parallel-Plate Air Capacitor?

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SUMMARY

The discussion centers on calculating the parameters of a parallel-plate air capacitor with plates measuring 16 cm², spaced 3.7 mm apart, and connected to a 12-V battery. The relevant equations include capacitance (C=KC₀=Kε₀A/d), charge (Q=CV), electric field (E=Q/(ε₀A)), and energy stored (U=0.5QV=0.5CV²). The calculated capacitance values varied, with one participant obtaining 3.83 pF and another 6.13 x 10^-11 F, both of which were deemed incorrect. The discussion highlights the importance of unit conversion and potential errors in the answer key of the educational platform used.

PREREQUISITES
  • Understanding of capacitance and its formulas
  • Familiarity with unit conversion, particularly from cm² to m²
  • Basic knowledge of electric fields and energy storage in capacitors
  • Experience with physics problem-solving in educational platforms like Mastering Physics
NEXT STEPS
  • Review the derivation of the capacitance formula for parallel-plate capacitors
  • Practice unit conversion techniques, especially for area measurements
  • Explore the implications of dielectric constants in capacitor calculations
  • Investigate common errors in physics problem-solving on educational platforms
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or capacitor design will benefit from this discussion.

thunderjolt
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A parallel-plate air capacitor is made by using two plates 16 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. What is the capacitance? What is the charge on each plate? What is the electric field between the plates? What is the energy stored in the capacitor?



What I believe to be relevant equations:
C=KC_0=Kε_0*A/d
Q=CV
E=Q/(ε_0*A)
U=.5QV=.5CV^2
K for air is1.00059




3. The Attempt at a Solution :
Really I'm just stuck on the first part. I tried C=KC_0=Kε_0*A/d, and put in C=(1.00059*8.854*10^-12*.0016/.0037)
 
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thunderjolt said:
C=(1.00059*8.854*10^-12*.0016/.0037)

Yes, evaluate.

ehild
 
Be careful with the area of the plates, particularly in converting to square meters.

Is "16 cm square" the same as 16 cm2, or does it mean (16 cm) x (16 cm)?
 
gneill said:
Be careful with the area of the plates, particularly in converting to square meters.

Is "16 cm square" the same as 16 cm2, or does it mean (16 cm) x (16 cm)?

That's the exact question word-for-word, so I can't tell either. The answer I got from doing the calculation was 3.83pF, which is wrong.
 
thunderjolt said:
That's the exact question word-for-word, so I can't tell either. The answer I got from doing the calculation was 3.83pF, which is wrong.

Suggest you try the other interpretation.
 
gneill said:
Suggest you try the other interpretation.

The other option also resulted in a wrong answer. 6.13*10^-11F using an area of (.16m)^2. I've 3 tries left.
 
thunderjolt said:
The other option also resulted in a wrong answer. 6.13*10^-11F using an area of (.16m)^2. I've 3 tries left.

Does the system want the answer in any particular units? Significant figures?
 
gneill said:
Does the system want the answer in any particular units? Significant figures?

It asks for Farads, and to put everything as XXX*10^Z. I have done so. I've used Mastering Physics all last year, so I know how to deal with the inputs.

Is there some special integral or another equation that I could try?

P.S. The way I'm doing the problem is exactly like the example in the book.
 
thunderjolt said:
It asks for Farads, and to put everything as XXX*10^Z. I have done so. I've used Mastering Physics all last year, so I know how to deal with the inputs.

Is there some special integral or another equation that I could try?

P.S. The way I'm doing the problem is exactly like the example in the book.

The equation is correct, and it appears to me that you've obtained a correct answer. If the method of data entry is also correct then it is possible that the machine's answer key is wrong. This has been known to happen from time to time. You should report it to your course instructor.

You might want to see if the capacitance value obtained returns correct results for other parts of the problem.
 

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