Parallel-plate capacitor question

[conniechiwa]

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!

 

[Defennder]

What's the right answer?

 

[alphysicist]

Hi conniechiwa,

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.

 

[conniechiwa]

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

 

[alphysicist]

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 $1.28\times 10^{-9}$ F, instead of the correct $1.128\times 10^{-9}$ 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 $3\times 10^{-6}$ microvolts is equal to $3\times 10^{-12}$ 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)

 

[conniechiwa]

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