<p>Does anyone has 2002 ap physics b form b solution? Thanks a lot</p>
<p>AP Physics B 2002 B Free Response Solution (I think)
(Problem 5)
a) Indicate on the diagram which plate is positive (+) and which is negative (-).
The plate on the left is negative and the plate on the right is positive. Negative fields attract inwards and positive fields attract outwards (in a sense). The arrows are pointing to the left, so that's how you know.</p>
<p>(b) Determine the potential difference between the plates.
V=Ed 100V = (5000N/c)*(2x10^-2 m)</p>
<p>(c) Determine the capacitance of this arrangement of plates.
C= eo(epsilon nought)*A/d = 8.85x10^-12 (0.3m2 /2x10-2m) = 1.33x10^-10 F</p>
<p>An electron is initially located at a point midway between the plates. </p>
<p>(d) Determine the magnitude of the electrostatic force on the electron at this location and state its direction.
F=Eq = (5000N/c)*(1.6x10-19 C) = 8 x10^-16N
Direction: to the left ( i think)</p>
<p>(e) If the electron is released from rest at this location midway between the plates, determine its speed just before striking one of the plates. Assume that gravitational effects are negligible.
V^2=Vo^2+2ax V=√2ax a=F/m v=√2xF/m v= √2 (8x10-16N)(0.1)/9.1x10-31) v= 4.2 x10^6 m/s</p>
<p>For any equations, look them up on the formula sheet, and just derive the proper ones (c'mon, basic algebra)</p>