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.?

Question

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?

az_lender · Answer

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.

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.?

Trending Questions