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Physics 7D Quiz 3 Name: Jordan Smolinsky Two flat plates with charge density σ = +5nC/m2 and −σ = −5nC/m2 , respectively, lie parallel to the xy plane, separated by a distance d = 2 cm. An electron (charge −1.6 × 10−19 C, mass 9.11 × 10−31 kg) enters the plates at height d/2 = 1 cm moving in the positive ŷ direction with speed 106 m/s. What is the electron’s speed when it hits the positively charged plate? Solution This problem can be approached in two ways: either by finding the force exerted by the plates’ electric field on the electron or by using conservation of energy. Since you have just covered potentials the preferred solution, which I will present here, is the latter. First we ~ = σ/0 ẑ want to find the potential between the plates. Because the electric field is constant E the potential difference between any two heights is simple to calculate. ~ · ~l = ∆V = E σ ∆z 0 (1) Therefore the potential difference between where the electron enters the plates and where it hits the positively charged plate is given by ∆V = σd 20 (2) eσd 20 (3) and the potential energy lost is then ∆U = − To find the final velocity we use conservation of energy. The final velocity is related to the initial velocity and the change in potential energy by 1 eσd 1 me vi2 + = me vf2 2 20 2 r eσd vf = vi2 + me 0 Plugging in our numbers we obtain 1.73 × 106 m/s. 1 (4)