How Do You Calculate the Potential Difference in a Parallel Plate Capacitor?

[Boozehound]

A parallel-plate capacitor has an area of 4.77 cm2, and the plates are separated by 1.06 mm with air between them. It stores a charge of 403 pC. What is the potential difference across the plates of the capacitor?

alright so i changed the area to .0477m^2. then i changed the distance between the plates to .00106m. i used 1.00054 as the dielectric constant. and i changed the charge to 403E-12C.

then i took 403E-12=((1.00054)(8.85E-12)(.0477)/(.00106))V

and i get V to equal 1.01V and its wrong. any help is appreciated.

 

[Dick]

For one thing, 4.77 cm^2 is 0.000477 m^2.

 

[m2003]

I would first like to commend your efforts in trying to solve this problem and using the appropriate units and constants. However, it seems that there may be a mistake in your calculation. The formula for calculating the potential difference across a parallel plate capacitor is V=Qd/εA, where Q is the charge, d is the distance between the plates, ε is the permittivity of the medium (in this case, air), and A is the area of the plates.

Using the values given in the problem, we can calculate the potential difference as:

V = (403E-12 C)(0.00106 m) / (8.85E-12 C^2/Nm^2)(0.0477 m^2)

V = 9.09 V

I believe the issue in your calculation may have been using the dielectric constant instead of the permittivity of air. The dielectric constant is only necessary when a dielectric material is present between the plates, but in this case, we are dealing with air as the medium.

I hope this helps. Keep up the good work in your scientific endeavors!