Question

Two 2.0-cm-diameter-disks spaced 1.5 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×10⁵ V/m .

Part C

An electron is launched from the negative plate. It strikes the positive plate at a speed of2.0×10⁷ m/s . What was the electron's speed as it left the negative plate?

Express your answer with the appropriate units.

Answer 1

1) given

diameter=2cm=0.02m

r=0.01m

d=1.5 mm=1.5*10^-3m

E=5.0×105 V/m

v=2*106/7m/sec

Since the electric field is constant inside the capacitor, the potential is just V = Ed, where d is the separation distance. So, V = Ed = (5 × 105 ) (.002) = 1000 V