Hi, I have a feeling something simmilar will be on my exam, so can you please gi

Question

Hi, I have a feeling something simmilar will be on my exam, so can you please give me the solutins and explain how to do this porblem. THANKS!

An electron traveling horizontally with a speed v0 = 1.5×107 m/s enters a uniform, vertical electric field created by two oppositely charged parallel plates as shown in Figure. The electric field between the plates has a magnitude of E = 2.0 ×104 N/C. The separation between the plates is s = 1.0 cm, and the vertical deflection of the electron as it passes between the plates is d = 0.60 cm.

1. What is the potential difference between the plates?

2. Which plate is at the higher voltage?

3. Through what potential difference does the electron travel as it passes between the plates?

4.  What is the speed of the electron as it leaves the area between the plates?

Explanation / Answer

1. V=Es = 2e04 N/C* (0.01m) = 200 Nm/C = 200V
2. Lower plate (positive plate)
3. V=Ed = 2e04 N/C * 0.006m = 120V
4. Horizontal component of the velocity of the electron as it leaves the two plates=v0=1.5e07 m/s
Vertical component can be found from
(If we do not neglect electron mass)
qE + mg = ma
a = (qE+mg)/m = (1.6e-19*2e04+9.1e-31*9.81)(9.1e-31) = 3.52e15 m/s2
v2 = u2+2as = 0 + 2(3.52e15)(0.006) = 6.5e6 m/s

Speed = (1.5e072+6.5e062) = 1.63 e07 m/s

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