Find the initial total charge stored

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SUMMARY

The discussion focuses on calculating the total charge stored in a parallel configuration of a 1.52 µF and a 2.63 µF capacitor with a voltage of 9.80 V. The initial total charge stored was correctly calculated as 4.067e-5 C. The challenge arises in determining the final total charge after reversing the connections of one capacitor, which requires understanding the effects of connecting capacitors in series. Additionally, the loss of electrical potential energy needs to be addressed, utilizing the formula C=AEoK/d, where Eo is permittivity and K is the dielectric constant.

PREREQUISITES
  • Understanding of capacitor configurations (parallel and series)
  • Knowledge of charge calculation in capacitors
  • Familiarity with electrical potential energy concepts
  • Basic grasp of dielectric materials and their properties
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  • Learn about calculating final charge in series capacitor configurations
  • Study the principles of electrical potential energy loss in capacitors
  • Explore the effects of reversing capacitor connections on charge distribution
  • Investigate the role of permittivity and dielectric constants in capacitor performance
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Electrical engineers, physics students, and anyone involved in circuit design or capacitor analysis will benefit from this discussion.

thst1003
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Initially a 1.52 µF capacitor and a 2.63 µF capacitor are connected in parallel, with 9.80 V across their plates. The two capacitors are not connected to anything else. The connecting wires are then reversed on the terminals of one capacitor, so that the positive plate of each capacitor is connected to the negative plate of the other capacitor.

a) Find the initial total charge stored.
I solved for it as if it was in parallel. I got 4.067e-5 C. This is correct.


(b) Find the final total charge stored.
I attempted to solve using it in series but I could not get the correct answer.


(c) Find the loss of electrical potential energy.


C=AEoK/d Eo represents Permitivvity and K is Kappa's constant of dialectrics



PLEASE PLEASE PLEASE HELP!
 
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hi thst1003! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)

if I'm understanding this correctly, you have charges +A and +B on the two connected plates on one side, and -A and -B on the other side

then you swap the wires so that +A and -B are connected (and -A and +B) …

so what happens to the charge? :wink:
 


Your 4.067E-5 number looks okay for the initial charge. You should also calculate the charge on the individual capacitors -- you'll need them later!

For part (b), one of the capacitors was inverted and reconnected. So if it had +++ charge on its top plate, and - - - on its bottom plate, that would now be reversed. What do you think will happen when the capacitors are connected in that orientation?
 

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