In the figure, a uniform, upward-pointing electric field E of magnitude 5.00Time

Question

In the figure, a uniform, upward-pointing electric field E of magnitude 5.00Times10^3 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 4 cm and separation d = 2.00 cm. Electrons are shot between the plates from the left edge of the lower plate. The first electron has the initial velocity v_0, which makes an angle 0=45degree with the lower plate and has a magnitude of 9.92Times10^6 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates. Another electron has an initial velocity which has the angle 0=45degree with the lower plate and has a magnitude of 8.01Times10^6 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates

Explanation / Answer

(a)   vertical velocity = 9.92 *106sin45

                                      = 7.01*106 m/sec

deacceleration due to electric field = 1.6*10-19*5*103/9.11*10-31

                                                         = 8.78*1014 m/sec2

Vertical distance travelled before velocity becomes zero = (7.01*106)2/2*8.78*1014

                                                                                                                          = 2.798 cm

This distance is greater than separation between plates which is equal to 2 cm

So, electron will hit the above plate.

Let after time t electron hit above plate so, vertical distance travelled by electron = 2cm

S = ut-1/2at2

0.02 =7.01*106t - 1/2*8.78*1014*t2

Solving for t

t = 0.38 *10-8 sec

Horizontal distance from left edge = 9.92 *106cos45 * 0.38 *10-8

                                                                                = 2.66 cm

(b) vertical velocity = 8.01 *106sin45

                                      = 5.66*106 m/sec

deacceleration due to electric field = 1.6*10-19*5*103/9.11*10-31

                                                         = 8.78*1014 m/sec2

Vertical distance travelled before velocity becomes zero = (5.66*106)2/2*8.78*1014

                                                                                                                          = 1.82 cm

As, this distance is less than 2 cm electron will not hit the plate .

=> Vertical distance at which particle leaves space between plates = 1.82 cm

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