Consider a horizontal square plate capacitor of area 1*1 m2, capacitance in vacu

Question

Consider a horizontal square plate capacitor of area 1*1 m2, capacitance in vacuum 3.5 F, which contains a dielectric material with dielectric constant K=5. The dielectric slides frictionlessly and is attached via a massless string and a massless pulley to a block of mass 2 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 1000 V

Explanation / Answer

So the energy of the capacitor has increased by: E1 = C x(U22-U12) / 2 = 1.68J Potential energy has decreased by E2 = mg h = 20 J So the kinetic energy of the block is K = E2 - E1 = 18.32 J So the velocity of the block is v = sqrt (2K/m) = 4.28 m/s

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