How would you modify this capacitor so that it can store 50J of energy without changing the charge of this plate
Answers
Answered by
7
Hi..
Q=CV. This relation has to be constant.
If dielectric happens to be air or fluid , and presuming plates are movable, if distance between plates us halved, C is doubled. The voltage becomes half of what it was originally.
Stored energy in capacitor is
E= 0.5 CV^2.
So now C is doubled, and V is half. Stored energy thus becomes half of original magnitude.The energy lost has been used up in bringing the plates together.
Reverse will happen if the distance is increased, when work will have to be done to separate the plates farther and this gets stored as energy added to capacitor.
Hope this helps u!!!
Q=CV. This relation has to be constant.
If dielectric happens to be air or fluid , and presuming plates are movable, if distance between plates us halved, C is doubled. The voltage becomes half of what it was originally.
Stored energy in capacitor is
E= 0.5 CV^2.
So now C is doubled, and V is half. Stored energy thus becomes half of original magnitude.The energy lost has been used up in bringing the plates together.
Reverse will happen if the distance is increased, when work will have to be done to separate the plates farther and this gets stored as energy added to capacitor.
Hope this helps u!!!
Answered by
6
Hey !!
Without changing the charge on parallel plates, the value of capacitance C will be half. From the equation, C = ε₀A/d if separation between the plates d becomes double, the capacitance C will become half while keeping charge Q constant.
Good luck !!
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