Week 1
Week 2
Solution. The electron is exerted an electric force toward the positive
plate. Find the force; F =
-eE = ma. The electron follows a projectile path, with acceleration down, a
= -eE/m. From the vertical motion: v_{y}^{2} = v_{y}_{0}^{2} + 2ad; To avoid the electron hitting the top plate v_{y}
must be zero. 0 = v_{y}_{0}^{2} + 2ad; (v_{0}sin45º)^{2} = 2eEd/m = 2(1.602
x 10^{-19 }C)(1.0 x 10^{4} N/C)(0.02 m)/(9.1 x 10^{-31}
kg), |
An electron beam is deflected upward through 3.0 mm while traveling in a vacuum between two deflection plates 12.0 mm apart (Figure). The potential difference between the deflecting plates is 100.0 kV and the kinetic energy of each electron as it enters the space between the plates is 2.0 x 10^{-15} J. What is the kinetic energy of each electron when it leaves the space between the plates?
Solution. Use conservation of energy, W_{field} = -ΔU, and the fact that the field is uniform to find K_{f}. ΔK = -ΔU K_{f} = K_{i} - ΔU = K_{i} + W_{field} = K_{i}+ eEΔy = K_{i}+ e(ΔV/d)Δy =
2.0 x 10^{-15} J + (1.602 x 10^{-19} C)(100.0 x 10^{3}
V)(0.0030 m)/0.0120 m) =
6.0 x 10^{-15} J |