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PHySics

A 1.63 nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00.
(a) How much work is required to withdraw the mica sheet?

Work = .5CV^2
C= k*E_0*A/d

But how can I solve for capacitance if I don't have area? And even if I find Capacitance how will I find work if I don't have volts?
THank You

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Physics

  1. When the dielctric material is withdrawn, the capacitance will decrease by a factor of 5, since it is proportional to the dielectric constant. Since C = Q/V and the charge Q remains the same, V will increase by a factor of 5.
    The stored capacitor energy, (1/2) CV^2, will increase by a factor (1/5)*5^2 = 5

    Initial stored energy
    E1 = (1/2)*1.63*10^-9*100^2 = 8.15*10^-6 J
    After withdrawing the mica,
    E2 = 4.08*10^-4 J

    The difference is the work required to remove the mica.

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    drwls

  2. I understand now, thank you.

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    Physics

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