Calculating Speed of Block with Dielectric in Capacitor

[123yt]

Homework Statement

Consider a horizontal square plate capacitor of area 1*1 m2, capacitance in vacuum 2 uF, 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 2.5 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 500 V.

Homework Equations

U = 1/2 CV^2 = q^2 / 2C

The Attempt at a Solution

Energy without dielectric = 1/2 * 2 * 10^-6 * 500^2 = 0.25
Energy with dielectric = 0.25 / k = 0.05

Energy change = 0.25 - 0.05 = 0.2

0.2 = 1/2 mv^2
v^2 = 0.4 / 2.5
v = 0.4 m/s

Actual answer is 4.4091 m/s .

 

[ehild]

Energy change = 0.25 - 0.05 = 0.2

Do not forget the change of the potential energy of the box.

ehild