Question 19
An electron travelling
horizontally in a vacuum enters the region between two horizontal
metal plates, as shown
in Fig. 6.1.
Fig. 6.1
The lower plate is
earthed and the upper plate is at a potential of + 400 V. The separation of the
plates is 0.80 cm.
The electric field
between the plates may be assumed to be uniform and outside the plates to be
zero.
(a)
On Fig. 6.1,
(i)
draw an arrow at P to show the direction of the force on the
electron due to the
electric field between
the plates,
(ii)
sketch the path of the electron as it passes between the plates and
beyond them.
[3]
(b)
Determine the electric field strength E between the plates. [2]
(c)
Calculate, for the electron between the plates, the magnitude of
(i)
the force on the electron,
(ii)
its acceleration.
[4]
(d)
State and explain the effect, if any, of this electric field on the
horizontal component of the motion of the electron. [2]
Reference: Past Exam Paper – November 2002 Paper 2 Q6
Solution:
(a)
(i)
{An electron
is negative and so, will be attracted towards the positive plate.}
arrow in
upward direction, foot near P
(ii)
Curved path consistent with (i) between plates then
straight (with no kink at change-over)
{The electron
moves horizontally and experiences a vertical force. This causes the electron
to move in a path similar to a projectile motion.
The electron
experiences a force only within the plates. Once it is outside the plates, the
path of the electron is straight.}
(b)
E = V / d
E = 400 /
(0.8×10-2) = 5.0×104
V m-1
(c)
(i)
F = Eq
F = (5.0×104) × 1.6×10-19 = 8.0×10-15
N
(ii)
{F = ma where m is the mass of an electron}
a = F / m
a = (8.0×10-15) / (9.1×10-31)
= 8.8×1015 m s-2
No comments:
Post a Comment
If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation