Work Done by Increasing Plate Distance of Flat Capacitor

In summary, the conversation discusses a flat capacitor with a distance of 2 cm and an area of 1 dm² that is connected to a voltage of 100 V. The voltage source is then disconnected and the distance between the plates is increased to twice the original distance. The conversation then goes on to calculate the capacitance at points A and B, the charge on the plates, the voltage of plate B, and the electrical energy in both plates. Finally, the conversation discusses the work done in this scenario, which is calculated to be 2.21 x 10^-8 J. The expert confirms that the calculations are correct but notes that the distance compared to the size of the plates may affect the accuracy of the capacitance calculation.
  • #1
mmoadi
157
0

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?:confused:
Thank you for helping!:smile:
 
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  • #2
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
:smile:Well, at least I cracked one problem!:biggrin:
Thank you for your help and HAPPY NEW YEAR:wink:!
 

1. What is the purpose of increasing the plate distance of a flat capacitor?

Increasing the plate distance of a flat capacitor allows for a larger electrical charge to be stored between the plates. This can increase the overall capacitance of the capacitor and allow it to store more energy.

2. How does increasing the plate distance affect the capacitance of a flat capacitor?

Increasing the plate distance of a flat capacitor increases the capacitance by creating a larger distance between the plates for the electrical charge to be stored. This also decreases the electric field strength between the plates.

3. Is there a limit to how much the plate distance can be increased in a flat capacitor?

There is no specific limit to how much the plate distance can be increased in a flat capacitor. However, as the distance between the plates increases, the capacitance also increases at a slower rate. Additionally, increasing the plate distance too much can lead to practical limitations such as a larger physical size for the capacitor.

4. What is the effect of increasing the plate distance on the energy stored in a flat capacitor?

Increasing the plate distance of a flat capacitor increases the energy stored by allowing for a larger electrical charge to be stored between the plates. This means that the capacitor can hold more energy and can be used for a longer time.

5. Are there any disadvantages to increasing the plate distance in a flat capacitor?

One potential disadvantage of increasing the plate distance in a flat capacitor is that it can lead to a decrease in the electric field strength between the plates. This can result in a lower rate of energy transfer and slower charging and discharging of the capacitor. Additionally, increasing the plate distance may also require a larger physical size for the capacitor, making it less practical for certain applications.

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