Consider a horizontal square plate capacitor of area 1.4*1.4 m2, capacitance in

Question

Consider a horizontal square plate capacitor of area 1.4*1.4 m2, capacitance in vacuum 2 µF, which contains a dielectric material with dielectric constant K=3. The dielectric slides frictionlessly and is attached via a massless string and a massless pulley to a block of mass 0.5 kg. The block pulls the dielectric from the capacitor as it falls. Compute the speed of the block at the instant the dielectric leaves the capacitor assuming it starts at rest and that the voltage across the capacitor after the dielectric is removed is measured to be 100 V.

Hint: Conserve energy to solve this problem. Remember that V and C are defined for the capacitor with no dielectric.

Explanation / Answer

Given Area of square plate capacitor = 1.4 x 1.4 m2 Capacitance in vacuum            = 2 F = 2 x 10-6F Dielectric constant                   = 3 mass of the block                      = 0.5 kg voltge (V)                                =100 V ----------------------------------------------------------------------------- Capacitance in vacuum            = 2 F = 2 x 10-6F Dielectric constant                   = 3 mass of the block                      = 0.5 kg voltge (V)                                =100 V ----------------------------------------------------------------------------- Energy of a capacitor without dielectric E ==U2C / 2
Energy of a capacitor with dielectric       E =U2C / 2K
Where U and C be the voltage and capacitance in vacuum of the capacitor.
From conservation of energy Energy of capacitor before pull the dielectric is equal to after pull by the block. Due to pull there exist potential ,as well as kinetic energy also to dielectric                U2C / 2K = - (m )(g) (l) + U2C / 2 + (1/2) mv2

(m )(g) (l) - U2C / 3 = (1/2) mv2

so                          v = (2 l g-2 U2C/3m) substitute given values, we get                                 v = (2 (1.4 m) (9.8 m/s2) - 2 (100V )2 (4 x 10-6 F / 3(0.5 kg)                                                       =5.238 m / s

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