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Unformatted text preview: Part b) We have already argued that the ﬁeld must point in the y direction, we must now
decide whether it points in the +ˆ or the −y direction. Recalling that ﬁeld lines originate on
positive charges and terminate on negative charges, it should be apparent that the ﬁeld points in
the −y direction.
Two large parallel copper plates are 5.0 cm apart and have a uniform electric ﬁeld between them
as depicted in ﬁgure 3. An electron is released from the negative plate at the same time that a
proton is released from the positive plate. Neglect the force of the particles on each other and
ﬁnd their distance from the positive plate when they pass each other. (Does it surprise you
that you need not know the electric ﬁeld to solve this problem?) Figure 3: Problem 22.51
This is really more of a mechanics problem than an electrodynamics one. All we really need to
recall from this chapter is that the force on a particle in the presence of an electric ﬁeld is given
F = qE
Clearly, both particles are going to experience the same force, but it will be in opposite directions.
The acceleration that each will experience is:
ae = −
Since the accelerations are constant and both particles are released from rest, we can express the
positions of both as functions of time.
xe = d −
The question that we are asking is thus, when does xe = xp = x0 . Setting both equal to x0 and
solving the system of equations gives:
x0 = d − q
me m p x0
qE ⇒ x0 = me
me + mp d = 27.3 µm This result is independent of the strength of the electric ﬁeld because the electron and the proton
have equal and opposite charge. 4
4.1 Chapter 23: Gauss’s Law
Flux Gauss’s law is a very powerful tool for solving highly symmetric problems. Before we can discuss
Gauss’s law though, we need to discuss the concept of ﬂux.
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