S.5 Physics (Electricity)

sap00acm

2010-11-09 4:40 am

The electric field strength between two parallel plates is 8 x 10^4 N/C and the separation between them is 4 cm. An electron accelerates from negative plate to the positive plate. If the initial speed is 2 m/s, what is the speed of the electron when it reaches the positive plate? (Given ε0 = 8.85 x 10^-12 C^2 / Nm^2, the mass of electron = 9.1 x 10^-31 kg)

回答 (1)

用戶25975

2010-11-09 5:56 am

✔ 最佳答案

According to V/d = E for parallel plate capacitor, we have, the potential difference between the plates equal to:

V = 8 x 10^4 x 0.04 = 3200 V

Thus, initial k.e. of the electron = 0.5 x 9.1 x 10^-31 x 2^2 = 1.82 x 10^-30 J

When the electron arrives at the positive plate, its gain in k.e. is:

1.6 x 10^-19 x 3200 = 5.12 x 10^-16 J

So its final k.e. is approx. equal to 5.12 x 10^-16 J since 5.12 x 10^-16 >> 1.82 x 10^-30

So speed of electron is given by:

0.5 x 9.1 x 10^-31 x v^2 = 5.12 x 10^-16

v = 3.35 x 10^7 m/s

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