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In part (a) of the figure an electron is shot directly away from a uniformly charged plastic sheet, at speed vs = 1.80 × 10^5 m/s.?
In part (a) of the figure an electron is shot directly away from a uniformly charged plastic sheet, at speed vs = 1.80 × 10^5m/s. The sheet is nonconducting, flat, and very large. Part (b) of the figure gives the electron's vertical velocity component v versus time t until the return to the launch point. What is the sheet's surface charge density?
1 Answer
- az_lenderLv 72 months ago
The sheet is positively charged, that's why the electron returns to the launch point. The magnitude of the acceleration is apparently (3.60 x 10^5 m/s)/(12 x10^(-12)s)
= 3.0 x 10^(16) m/s^2.
The electric field in this area is
sigma/(2*epsilon-nought), and must also equal ma/Q,
where m and Q are the charge and mass of the electron. So sigma is your only unknown, and is 2ma*(epsilon-nought)/Q, where the "a" comes from my first paragraph.