Q: Two plates are placed 0.1m apart with a potential difference of 20V. An electron is initially at rest on the surface of the plate with the lower potential. What will be the velocity of the electron when it strikes the other plate?

A 1.87*10^6 m/s
B 2.65*10^6 m/s
C 7.02*10^12 m/s
D 3.2*10^-19 m/s
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Answer: B

Explanation:

The potential difference between the plates is equal to the change in electric potential energy of the electron. The potential difference is given by V = Ed, where E is the electric field intensity between the plates and d is the distance between them. The electric field intensity E = V/d = 20V/0.1m = 200 V/m. The change in electric potential energy is qV, where q is the charge of the electron (-1.6*10^-19 C). Setting the change in potential energy equal to the change in kinetic energy (0.5mv^2), we have qV = 0.5mv^2. Solving for v, we get v = sqrt(2qV/m) = sqrt(2*(-1.6*10^-19 C)*(20V)/(9.11*10^-31 kg)) ≈ 2.65*10^6 m/s.

Q: Two plates are placed 0.1m apart with a potential difference of 20V. An electron is initially at rest on the surface of the plate with the lower potential. What will be the velocity of the electron when it strikes the other plate?

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