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Capacitor - with and without dielectric

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(1) A parallel plate capacitor of plate area 2m2 and palte separation 5mm is charged to 10,000V in vacuum. Compute the capacitance, charge, charge density, field strength in the space between the plates and also the energy stored in the capacitor.

(2) The charging battery is removed and the space between the plates is filled with a material of dielectric constant 5. Compute the new capacitance, the potential difference, the field strength and the energy stored.

(3) The charging battery remains connected and the space between the plates is filled with a material of dielectric constant 5. Compute the new capacitance, the charge, the field strength and the energy stored.

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Detailed explanations on the effect of dielectric on the capacitance and in the stored charge with answers. Answer as pdf attachment.

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The capacitance,
Cair = "0A
d
Given A = 2m2, d = 5mm = 5*10−3m
Therefore Cair = C0 =8.8510−122
510−3 ...

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