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

September 28th, 2009 | by Michael |
energy
lildrak2002 asked:


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?

BENNIE

  1. 2 Responses to “How much energy is stored in the capacitor before the dielectric is inserted?”

  2. By andyg77 on Sep 29, 2009 | Reply

    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.

  3. By ukmudgal on Oct 2, 2009 | Reply

    ______________________________
    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
    ________________________
    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
    __________________________________
    Part C
    Energy increases by 12.46375 *10^-3 J during the insertion.
    ____________________________________

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