Two 2.5-cm-diameter-disks spaced 1.5 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.5×105 V/m .
An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.5×107 m/s . What was the electron's speed as it left the negative plate?
V = E*d = 4.5*10^5 * 1.5*10^-3 =675 Volt
By energy conservation suppose electron speed as it leaves the -ve plate =vi
1/2*m*vi^2 = qΔV + 1/2*m*vf^2
charge of the electron q = -1.6*10^-19 C ,
and its mass m = 9.11*10^-31 kg
0.5*9.11*10^-31*vi^2 = -1.6*10^-19*675 + 1/2*9.11*10^-31* ( 2.5*10^7)^2
v≈1.96951×10^7 m/s answer
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