A) A proton has an initial velocity of 2.79 × 107 m/s in the horizontal directio

Question

A) A proton has an initial velocity of
2.79 × 107 m/s in the horizontal direction.
It enters a uniform electric field of 3400 N/C
directed vertically.
Ignoring gravitational effects, find the
time it takes the proton to travel 0.096 m
horizontally. The mass of the proton
is 1.6726 × 1027 kg and the fundamental
charge is 1.602 × 1019 C .
Answer in units of ns.

B) What is the vertical displacement of the proton
after the electric field acts on it for that
time?
Answer in units of mm.

C) What is the proton’s speed after being in the
electric field for that time?
Answer in units of km/s.

Explanation / Answer

A.

The horizontal motion is unaffected by the vertical E field. Thus, it is just

t = dx / vx

= 0.096 m / 2.79E7 m/s

= 3.44E-9 s

= 3.44 ns [ANSWER]

**************************

B.

The acceleration of the proton is

ay = F / m = E q / m = 3.256E11 m/s^2

Thus,

y = voy t + 1/2 ay t^2

As voy = 0,

y = 1.93E-6 m

= 0.00193 mm [ANSWER]

**************************

C)

As

vy = voy + ay t

Then vy = 1120 m/s

As vx = 2.79E7 m/s, using Pythagorean theorem,

v = 2.79*10^7 m/s

=2.79*10^4 km/s [the velocity along y is so small that it didn't matter to 3 decimal places.]

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