How much energy is stored in the capacitor before the dielectric is inserted?

A 12.5 uF capacitor is connected to a power supply that keeps a constant potential difference of 26.0 V across the plates. A piece of material having a dielectric constant of 3.95 is placed between the plates, completely filling the space between them.



A)How much energy is stored in the capacitor before the dielectric is inserted?



Part B

How much energy is stored in the capacitor after the dielectric is inserted?



Part C -

By how much did the energy change during the insertion?How much energy is stored in the capacitor before the dielectric is inserted?
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Capacity = C = 12.5 uF 1.25*10^ - 5 F



Potential difference = V = 26.0 V across the plates.



Energy stored in capacitor before dielectric is placed = Ui = (1/2)CV^2



Ui = 0.5*1.25*10^ -5*26*26



Part A

Energy stored in capacitor before dielectric is placed =Ui =4.225*10^-3 J

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As potential is kept constant,insertion of dielectric increases capacity



Energy stored in capacitor after dielectric is placed =Uf =(diectric constant)CV^2 = (diectric constant)Ui= 3.95*4.225*10^-3



Part B

Energy stored in capacitor after dielectric is placed =Uf =1.668875*10^-2 J

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Part C

Energy increases by 12.46375 *10^-3 J during the insertion.

____________________________________How much energy is stored in the capacitor before the dielectric is inserted?
1.00059 = dielectric constant of air at 1 atm



The energy storage in a capacitor is given by 1/2*C*V^2



So just multiply it out (.5)*(.0000125Farads)*(26)^2 = .004225 joules.



Inserting the 3.95 dielctric constant material will increase the capacitance by the ratio 3.95/1.00059 = 3.9477



Since this is linear relationship just multiply (.004225)*3.9477= .0167 joules



So the stored energy increased from .004225 joules to .0167 joules and you can express that as you see fit, either as a percentage or as additional joules of energy.

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