Calculating Work Required to Remove Dielectric Slab from Charged Capacitor

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In summary, the parallel-plate capacitor initially has a capacitance of 2.19nF and is charged to an initial potential difference of 94V. The dielectric slab with a dielectric constant of 7.54 is then removed, leading to a change in capacitance and potential difference. The work required for removing the dielectric slab can be calculated using the formula W= Ui - Uf, where Ui is the initial potential energy and Uf is the final potential energy.
  • #1
srhly
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An isolated capacitor has a dielectric slab k between its two plates. The capacitor is charged by a battery. After the capacitor is charged, the battery is removed. The dielectric slab is then removed. Finally, the capacitor reaches equilibrium. Initially the parallel-plate capacitor has a capacitance of 2.19nF is charged to an initial potential difference of 94 V. The dielectric material has a dielectric constant k = 7.54. What is the magnitude of the work required when the dielectric slab is then removed?

I used:
W= Ui - Uf
U= 0.5CV^2
Uf= (0.5)C(k*Vi)^2
Ui = (0.5)C(Vi)^2

Vi=initial potential difference.

I have tried to double check the math, so I guess I messed up with the formulas. Where did I go wrong?
 
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  • #2
After charging, the battery is removed hence by displacing the slab the capacitance will change produces change in potential difference between the plates.

In such processes where the battery is disconnected the charge will remain unaltered, not potential difference.
 
  • #3


There appears to be an error in the formula used to calculate the final potential energy after the dielectric slab is removed. The correct formula should be Uf = (0.5)(C/k)(Vi)^2, as the dielectric material increases the capacitance by a factor of k. Using this formula, the final potential energy would be (0.5)(2.19nF/7.54)(94V)^2 = 6.44mJ.

The work required to remove the dielectric slab can then be calculated as W = Ui - Uf = (0.5)(2.19nF)(94V)^2 - 6.44mJ = 0.97mJ. This means that approximately 0.97mJ of work is required to remove the dielectric slab from the charged capacitor.

It is important to note that this calculation assumes ideal conditions and does not take into account any losses or inefficiencies in the system. In a real-world scenario, the work required may be slightly different due to these factors. Additionally, the exact method of removing the dielectric slab may also impact the amount of work required. Overall, this calculation provides an estimate of the work required and can be used as a starting point for further analysis and experimentation.
 

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