How Do You Calculate Electric Field and Potential in a Parallel Plate Capacitor?

[ledwardz]

Homework Statement

A parallel plate capacitor has a charge of 50pC
area = 10mm^2
dielectric material permittivity = 2.5
distance between plates = 0.5 mm

find the electric field
the potential difference between the 2 plates
the energy stored in the capacitor

Homework Equations

electric field = Charge density / permittivity
charge density = q / a

therefore E = 5*10^-3 / 2.5 = 2* 10 ^-3

PD= Ed = 2*10^-3 * 0.005 = 1*10^-5

energy u = 0.5V*q = 0.5 * 1*10^-5* 50*10^-9 = 2.5*10^-13 J

can anyone tell me if this is correct? thanks again for any help. Cheers, Lee:shy:

 

[gneill]

You seem to have lost track of the orders of magnitude in the charge density. Check the calculation.

The permittivity κ is a multiplier for the vacuum permittivity, ε_o.

 

[ledwardz]

You seem to have lost track of the orders of magnitude in the charge density. Check the calculation.

The permittivity κ is a multiplier for the vacuum permittivity, ε_o.

ahh pooo. so ε = ε_o*ε_r but i have the right idea for the rest of the equation?

thanks for reply also.

 

[gneill]

ahh pooo. so ε = ε_o*ε_r but i have the right idea for the rest of the equation?

thanks for reply also.

Sure. Straighten out the powers of ten and the permittivity and you're golden.

 

[rwisz]

Your calculations are correct. The electric field between the plates of the parallel plate capacitor is 2*10^-3 N/C, the potential difference between the plates is 1*10^-5 V, and the energy stored in the capacitor is 2.5*10^-13 J. These values are in accordance with the given charge, area, distance, and permittivity. Great job!