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Amount of work that is done to adjust the capacitor from position A to position B 
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#1
Dec2909, 12:02 PM

P: 157

_{1. The problem statement, all variables and given/known data
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?
2. Relevant equations
C= ε*A/ d
Q= C*V
E= ½ Q*V
W= ΔE
3. 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!}



#2
Dec2909, 05:52 PM

P: 1,398

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


#3
Dec2909, 06:30 PM

P: 157

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



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