Item 8 suppose an electron traveling with speed vo 1.15x10 m/s enters a uniform

Question

Item 8 suppose an electron traveling with speed vo 1.15x10 m/s enters a uniform electric feld E, which is at right angles to vo as shown in the figure (Figure 1) 1 of 1 0 vo Part A At what angle will the electrons leave the uniform electric field at the end of the parallel plates (point P in the figure)? Assume the plates are 52 cm long and E-5.5x103 N/O ignore fringing of the field. Express your answer to two significant figures and include the appropriate units. Value Units clockwise from tz axis My Answers Give Up Continue

Explanation / Answer

I'll use the notation 1.00E7 instead of 1.15x10^7 as it's easier and neater.

The only force is the electric force acting vertically downwards (because the electrons are negative).

F = qE = -1.6x10E-19 x 5500 = -8.80E-16 N (negative meaning downwards, in the -y direction)

a = F/m = -8.80E-16/9.11E-31 = -9.6597E14 m/s²

(The above is equivalent to using a = qE/m)

The x component of velocity, Vx, is constant: Vx = v = 1.15E7 m/s because there are no forces acting in the x direction. So we use this to get the time to pass through the plates
T = distance/speed = 0.052/1.15E7 = 4.52E-9 s

On entering, the y component of velocity is 0.
The y component of velocity on leaving is found by using the y acceleration:
Vy = at = -9.6597E14 x 4.52E-9 = 4.368E6 m/s

The angle is given by
tan = Vy/Vx = -4.368E6/1.15E7 = -0.3798
= tan¹(-0.3798) = -20.8

part b = 45 degree

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