Question

Two 1.60-cm-diameter disks spaced 1.50 mm apart from a parallel-plate capacitor. The electric field between the disks is 5.20 times 10^5 V/m. What is the voltage across the capacitor? Express your answer to two significant figures and include the appropriate units. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.40 times 10^7 m/s. What was the electron's speed as it left the negative plate? Express your answer to two significant figures and include the appropriate units.

Answer 1

The separation ofthe disks d = 1 .5mmFl.5x 10 3m The electric field E-5.2x10 V/m (A) The voltage across the capacitor is V 7x d = 5.2 x 10° × 1 .5x 103-780 volt (B) The force on the electron F -Eq Eq where q 1.6x10-1 C and 9.11x10-3 kg, are charge and mass of the electron 52x10' x1.6x10-9.13x10"ms The acelleration ofthe electron a 9.1 1 × 10-31 Here the final velocity v,-24 × 107 m / s Using v =V2.4x 107 )2-2×9. I 3 x1016×1.5x103 =V5.76 × 10' -2.73 × 10 =V3.03 x1014 = 1.74× 107ms So this is the electron speed at the negative plate.