There is given an ideal capacitor. The capacitor consists of two plates at a distance of 1mm, in a Cartesian coordinate system the plates are parallel to the (x,y) plane.

The capacitor will be charged such that an electrical energy of 10~ J will be stored in the capacitor and then it will be

disconnected from the voltage source. You know that the stored energy W of a capacitor is given by W = -QU where Q is

the stored charge and U is the applied voltage.

Now you bring in a dielectric material with a relative dielectric constant of 3 which exactly fits between the two plates. You bring in the dielectric material by moving it without friction along the x-axis. The dielectric material has a mass of 10g.

What will be the velocity Up of the dielectric material just at that moment when it is completely between the two plates of the capacitor?

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1. VD = 245 mm/s 2. VD 46,4 mm/s 3. VD = 3650 μm/s 4. VD = 355 μm/s 5. VD 7,81 . 10^-3 m/s = 6. VD= 4,52 mm/s 7. VD 7,96 10^-3 m/s 8. VD2,31. 10^-3 m/s 9. VD 88,2. 10^-3 m/s 10. VD = 26,3 . 10^-3 m/s = = + 1. VD = 245 mm/s 2. VD 46,4 mm/s 3. VD = 3650 μm/s 4. VD = 355 μm/s 5. VD 7,81 . 10^-3 m/s = 6. VD= 4,52 mm/s 7. VD 7,96 10^-3 m/s 8. VD2,31. 10^-3 m/s 9. VD 88,2. 10^-3 m/s 10. VD = 26,3 . 10^-3 m/s = = +

Answer rating: 100% (QA)

SOLUTION To determine the velocity of the dielectric material we need to use the equation for the stored energy of a capacitor which is W QU where Q is the stored charge and U is the applied voltage S... View the full answer