Q#19 (Past Exam Paper – November 2002 Paper 2 Q6)
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]
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
(d) There would be no effect as the electric force is normal to the horizontal component of the motion.
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