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

AI Thread Summary
To calculate the parameters of a parallel-plate air capacitor with 16 cm square plates spaced 3.7 mm apart and connected to a 12-V battery, the relevant equations include capacitance (C), charge (Q), electric field (E), and energy (U). The capacitance was initially calculated using C=KC_0=Kε_0*A/d, but there were concerns about the area conversion from cm² to m². Despite attempts to evaluate the capacitance, the results were inconsistent, leading to confusion about the correct interpretation of the plate area. It was suggested that if the calculations and data entry were correct, the issue might lie with the answer key provided by the course system. The discussion emphasized the importance of verifying the capacitance value against other parameters of the problem.
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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