Solving a Parallel-Plate Capacitor Problem

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In summary, the conversation is about a problem involving calculating the capacitance of a parallel-plate capacitor with circular plates. The formula for capacitance is discussed and the individual attempts to solve the problem using it but realizes that the area of a circle is incorrect. Once the mistake is pointed out, the problem is solved.
  • #1
bbbbbev
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Hi. I have this seemingly easy problem that I cannot get the right answer to! Here it is:

A parallel-plate capacitor has circular plates of 6.00 cm radius and 2.00 mm separation.
Calculate the capacitance.

Ok, so I thought the formula for capacitance was [(epsilon)_0 x Area]/distance = capacitance

So I tried: [(8.85E-12 F/m)(4(pi)(0.06m)^2)]/0.002m = 200.2 pF

I cannot figure out what I am doing wrong. That formula is in the book! Am I overlooking something? Thanks in advance for any help you can give!

Beverly
 
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  • #2
The area of a circle isn't 4*pi*r^2. :)
 
  • #3
Aggh! I'm such an idiot! What was I thinking? Haha, Thanks!
 

Related to Solving a Parallel-Plate Capacitor Problem

1. How do you calculate the capacitance of a parallel-plate capacitor?

The capacitance of a parallel-plate capacitor can be calculated using the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

2. What is the electric field between the plates of a parallel-plate capacitor?

The electric field between the plates of a parallel-plate capacitor can be calculated using the formula E = V/d, where V is the potential difference between the plates and d is the distance between the plates.

3. How does the distance between the plates affect the capacitance of a parallel-plate capacitor?

The capacitance of a parallel-plate capacitor is inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the capacitance decreases, and vice versa.

4. Can you use a parallel-plate capacitor to store energy?

Yes, a parallel-plate capacitor can store energy in the form of electrical potential energy. The amount of energy stored is equal to 1/2CV², where C is the capacitance and V is the potential difference between the plates.

5. How do you solve a parallel-plate capacitor problem involving dielectric materials?

To solve a parallel-plate capacitor problem involving dielectric materials, you will need to use the formula C = κε₀A/d, where κ is the dielectric constant of the material. You will also need to take into account the effect of the dielectric material on the electric field and potential difference between the plates.

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