Work Done by Increasing Plate Distance of Flat Capacitor

[mmoadi]

Homework Statement

Flat capacitor with a plates distanced d = 2 cm and the area A= 1 dm² are connected to a voltage U = 100 V. We disconnect the source of voltage and we increase the distance between the plates to twice the distance. How much work is done?

Homework Equations

C= ε*A/ d
Q= C*V
E= ½ Q*V
W= ΔE

The Attempt at a Solution

Calculating the capacitance at point A:
d(A)= 0.02 m, A= 0.01 m², ε= 8.85 × 10^-12 As/Vm

C(A)= ε*A/ d(A)
C(A)= [(8.85 x 10^-12 As/Vm)*(1 x 10^-2 m^2)]/ (2 x 10^-2 m)
C(A)= 4.43 x 10^-12 F

Calculating the capacitance at point B:
d(B)= 0.04 m, A= 0.01 m², ε= 8.85 × 10^-12 As/Vm

C(B)= ε*A/ d(B)
C(B)= [(8.85 x 10^-12 As/Vm)*(1 x 10^-2 m^2)]/ (4 x 10^-2 m)
C(B)= 2.21 x 10^-12 F

Calculating the charge on the plates:

Q= C(A)*V(A)
Q= (4.43 x10^-12)*100 V
Q= 4.43 x 10^-10 C

We know that the charge should be the same on both plate A and plate B so we use that piece of information to calculate the voltage of plate B:

V(B)= Q/ C(B)
V(B)= (4.43 x 10^-10 C)/ (2.21 x 10^-12 F)
V(B)= 200 V

Calculating the electrical energy in plate A:

E(A)= ½ Q*V(A)
E(A)= ½ (4.43 x10^10 C)* 100 V
E(A)= 2.22 x10^-8 J

Calculating the electrical energy in plate B:

E(B)= ½ Q*V(B)
E(B)= ½ (4.43 x10^10 C)* 200 V
E(B)= 4.43 x10^-8 J

Calculating the work:

W= ΔE
ΔE= E(B) – E(A)
ΔE= 4.43 x10^-8 J - 2.22 x10^-8 J
ΔE= 2.21 x 10^-8 J

Are my calculations correct?
Thank you for helping!

 

[willem2]

Your calculations are correct. The distance of the plates compared to their size is too large
to make the capacitance calculation so accurate however.

 

[mmoadi]

_{Well, at least I cracked one problem!
Thank you for your help and HAPPY NEW YEAR!}