Question

A proton has an initial velocity of 7.6 × 106 m/s in the horizontal direction. It enters a uniform electric field of 8800 N/C directed vertically.

a) Ignoring gravitational effects, find the time it takes the proton to travel 0.127 m horizontally. The mass of the proton is 1.6726 × 10−27 kg and the fundamental charge is 1.602 × 10−19 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.

Answer 1

Part A.

Since electric field is directed upward, So there will be zero force on proton in horizontal direction, And Since Force is zero, So acceleration of proton in horizontal direction will also be zero.

Which means

distance = Speed*time

time = distance/Speed

t = d/Vx

Using given values:

Vx = Horizontal speed = 7.6*10^6 m/sec

d = 0.127 m

t = 0.127/(7.6*10^6) = 1.67*10^-8 sec

t = 16.7*10^-9 sec = 16.7 ns

Part B.

Using Force balance in vertical direction

Fe = Fnet

q*E = m*ay

ay = acceleration in vertical direction = q*E/m

ay = 1.602*10^-19*8800/(1.6726*10^-27)

ay = 8.43*10^11 m/sec^2

Now Using 2nd kinematic equation in vertical direction

d = Vy*t + 0.5*ay*t^2

d = 0*16.7*10^-9 + 0.5*8.43*10^11*(16.7*10^-9)^2

d = 0.12*10^-3 m

d = 0.12 mm

Part C.

Since there is no acceleration in horizontal direction, So horizontal speed will remain constant

After time t = 16.7 ns

V1x = 7.6*10^6 m/sec

In vertical direction using 1st kinematic equation

V1y = Vy + ay*t

V1y = 0 + 8.43*10^11*16.7*10^-9

V1y = 14078.1 m/sec

Net velocity will be

|V1| = sqrt (V1x^2 + V1y^2)

|V1| = sqrt ((7.6*10^6)^2 + (14078.1)^2)

|V1| = 7600013 m/sec = 7600.013*10^3 m/sec

|V1| = 7600.013 km/sec

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