V= (1/eo) Q d / A
V = 828 I get
eo = 8.85*10^-12
d = 1.8*10^-3
A = pi r^2 = pi (1.15*10^-2)^2
Ke leaving plate = (1/2) m v^2
Ke gained = electron charge * Voltage
total Ke= (1/2) m (10^7)^2
Two 2.3-cm-diameter-disks spaced 1.8 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.6×105 V/m.
A. What is the voltage across the capacitor = 830 V (solved)
B. How much charge is on each disk? (Find Q1 and Q2)
C. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.3×107 m/s . What was the electron's speed as it left the negative plate?
I just need B and C. Thank you!!
1 answer
1 answer