PLEASE HELP WITH THESE TWO QUESTIONS! PLEASE HELP ME BY SHOWING THE STEPS! THANK

Question

PLEASE HELP WITH THESE TWO QUESTIONS! PLEASE HELP ME BY SHOWING THE STEPS! THANK YOU SO MUCH

An electron with a speed of 3.00 × 106 m/s moves into a uniform electric field of magnitude 1.00 × 103 N/C. The field lines are parallel to the electron’s velocity and pointing in the same direction as the velocity. How far does the electron travel before it is brought to rest?

(a) 2.56 cm (b) 5.12 cm (c) 11.2 cm (d) 3.34 cm (e) 4.24 cm

On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet’s surface. A 2.00-kg ball having a charge of 5.00 C is thrown upward at a speed of 20.1 m/s. It hits the ground after an interval of 2.40 s. What is the potential difference between the starting point and the top point of the trajectory?

Explanation / Answer

1)

To answer this question you find the electrostatic force (F) on the electron, then find its acceleration (a), and then use the formula for uniformly accelerated motion v^2 - u^2 = 2*a*d

F = q*E where q = electronic charge, E = electric field strength

a = F/m = q*E/m where m = mass of electron.

d = u^2/(2*a) = m*u^2/(2*q*E) where u is the original speed of the particle.

put = m = 9.10 × 10-31

speed u =3.00 × 106 m/s

E = 1.00 × 103 N/C.

q = 1.602 x 10-19c

= (a) 2.56 cm

2)

If you don't...

E=(m/q)(g-[2v0/t])

=(2.00kg/5.00e-6C)(9.8-[2*20.1m/s/2.40s])

=-5.7e3 N/C downward

v^2=v0^2+2ax

Becomes x=(0-v0^2)/(2a)

Becomes x=(v0t)/4

=(20.1m/s*2.40s)/4

=12.06m

delta V=-E*x
=-(-5.7e3 N/C * 12.06m)
=6.87e4 V
=68.7kV

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