1 answer

A proton moves at 5.10 x 10^5 m/s in the horizontal direction. It enters a uniform...

Question:

A proton moves at 5.10 x 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.10 x10^3 N/C. Ignore any gravitational effects.

(a) Find the time interval required for the proton to travel 4.70 cm horizontally.

(b) Find its vertical displacement during the time interval in which it travels 4.70 cm horizontally.

(c) Find the horizontal and vertical components of its velocity after it has traveled 4.70 cm horizontally.


Answers

a) Since the field is vertical, the only force is vertical and the horizontal velocity does not change. Therefore t =d/vh = 0.047/(5.10*10^5) s

= 9.22*10^-8 s

b) the vertical displacement in time t is 0.5*a*t²; t you have from part a), and acceleration

a = F/m = q*E/mp ; where q = elemental charge, mp = proton mass

= 7.76*10^11

So,

d = 0.5*(7.76*10^11)*( 9.22*10^-8)^2

= 3.3*10^-3 m

c)

Vh = 5.10*10^5 m/s

v = a*t, a and t as computed above.

Vv = at = 7.15*10^4 m/s

.

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