Parallel-plate capacitor question

  • Thread starter conniechiwa
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In summary: It's difficult to tell what you did without any details. However, I don't believe the capacitance you calculated is correct. (Perhaps you misread the result on your calculator?) Try it again, and if you don't get the right answer, please post some more details of the numbers you used.The answer I submitted for capacitance was accepted. This is what I did:Area = (50000 micrometers)(51000 micrometers) = 2.55E9 micrometersC=(E0A)/d =(8.85E-12)(2.55E9 micrometers)/20 micrometers
  • #1
conniechiwa
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A parallel-plate capacitor is made from two aluminum foil sheets, each 5 cm long and 5.1 cm wide. The two sheets are separated by a distance of 0.02 mm. If a potential difference of 3 volts is applied across the two plates of this capacitor, what is the absolute value of the charge on the plate with the higher voltage? What is the absolute value of the charge on the plate with the lower voltage?

I have already found the capacitance by using C = E0A/d. I got an answer of C = 1.28E-9 F. I tried using the equation Q=CV in order to find the value of the charge on the plates, but I still can't get the right answer. Help please!
 
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  • #2


What's the right answer?
 
  • #3


Hi conniechiwa,

conniechiwa said:
A parallel-plate capacitor is made from two aluminum foil sheets, each 5 cm long and 5.1 cm wide. The two sheets are separated by a distance of 0.02 mm. If a potential difference of 3 volts is applied across the two plates of this capacitor, what is the absolute value of the charge on the plate with the higher voltage? What is the absolute value of the charge on the plate with the lower voltage?

I have already found the capacitance by using C = E0A/d. I got an answer of C = 1.28E-9 F. I tried using the equation Q=CV in order to find the value of the charge on the plates, but I still can't get the right answer. Help please!

It's difficult to tell what you did without any details. However, I don't believe the capacitance you calculated is correct. (Perhaps you misread the result on your calculator?) Try it again, and if you don't get the right answer, please post some more details of the numbers you used.
 
  • #4


The answer I submitted for capacitance was accepted. This is what I did:
Area = (50000 micrometers)(51000 micrometers)
= 2.55E9 micrometers
C=(E0A)/d
=(8.85E-12)(2.55E9 micrometers)/20 micrometers
=1.128E-3 microFarads

To find charge, I did:
q=cv
=(1.128E-3 microFarads)(3E-6 microvolts)
= 3.84E-9 microcoulombs
 
Last edited:
  • #5


conniechiwa said:
The answer I submitted for capacitance was accepted. This is what I did:
Area = (50000 micrometers)(51000 micrometers)
= 2.55E9 micrometers
C=(E0A)/d
=(8.85E-12)(2.55E9 micrometers)/20 micrometers
=1.128E-3 microFarads

Right; that's what I got for the capacitance. (In your original post you said you got [itex]1.28\times 10^{-9}[/itex] F, instead of the correct [itex]1.128\times 10^{-9}[/itex] F, but maybe that was just a typo in your post?)

To find charge, I did:
q=cv
=(1.128E-3 microFarads)(3E-6 microvolts)
= 3.84E-9 microcoulombs

These numbers don't look right for the voltage. The [itex]3\times 10^{-6}[/itex] microvolts is equal to [itex]3\times 10^{-12}[/itex] volts, which is not right.

Also, for the units, microFarads times microvolts does not give microcoulombs; it would give something like micromicrocoulombs (since there are two "micro"s in the product), which you would definitely have to adjust.

In other words, the units would act like:

(1 microfarad) (1 microvolt) = (1e-6 microcoulomb) = (1e-12 coulomb)
 
  • #6


Oh okay I figured out what my problem was. Thanks. =)
 

1. What is a parallel-plate capacitor?

A parallel-plate capacitor is a device used to store electrical energy. It consists of two parallel conducting plates separated by a small distance and is typically used in electronic circuits.

2. How does a parallel-plate capacitor work?

A parallel-plate capacitor works by creating an electric field between the two plates. When a voltage is applied to the plates, one plate becomes positively charged while the other becomes negatively charged. The electric field between the plates stores the potential energy, which can be released when the capacitor is connected to a circuit.

3. What factors affect the capacitance of a parallel-plate capacitor?

The capacitance of a parallel-plate capacitor is affected by the distance between the plates, the surface area of the plates, and the dielectric material between the plates. A larger distance between the plates or a smaller surface area will result in a lower capacitance, while a higher dielectric constant material will result in a higher capacitance.

4. How is the capacitance of a parallel-plate capacitor calculated?

The capacitance of a parallel-plate capacitor can be calculated using the formula C = ε0A/d, where C is the capacitance in Farads, ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the surface area of the plates in square meters, and d is the distance between the plates in meters.

5. What are some practical applications of parallel-plate capacitors?

Parallel-plate capacitors are commonly used in electronic circuits for energy storage and filtering. They are also used in power factor correction and as sensors in touchscreens and accelerometers. Additionally, parallel-plate capacitors are used in energy storage systems for renewable energy sources such as solar panels and wind turbines.

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