(adsbygoogle = window.adsbygoogle || []).push({}); _{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!}

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# Homework Help: Amount of work that is done to adjust the capacitor from position A to position B

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