A proton moves at 4.80 105 m/s in the horizontal direction. It enters a uniform

Question

A proton moves at 4.80 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.40 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 5.50 cm horizontally. ns (b) Find its vertical displacement during the time interval in which it travels 5.50 cm horizontally. (Indicate direction with the sign of your answer.) mm (c) Find the horizontal and vertical components of its velocity after it has traveled 5.50 cm horizontally. v with arrow = î + km/s

Explanation / Answer

a)

v = speed of proton = 4.80 x 105 m/s

X = horizontal distance travelled = 5.50 cm = 0.055 m

t = time of travel = ?

using the equation

t = X/v

t = 0.055/(4.80 x 105)

t = 1.15 x 10-7 sec = 115 ns

b)

m = mass of proton = 1.67 x 10-27 kg

q = charge on proton = 1.67 x 10-19 C

E = electric field = 9400 N/C

acceleration is given as

a = qE/m

a = (1.67 x 10-19) (9400)/(1.67 x 10-27)

a = 9.4 x 1011 m/s2

vo = initial velocity in vertical direction = 0 m/s

t = time of travel = 1.15 x 10-7 sec

Y = vertical displacement

vertical displacement is given as

Y = vo t + (0.5) a t2

Y = 0 (1.15 x 10-7) + (0.5) (9.4 x 1011) (1.15 x 10-7)2

Y = 0.00622 m = 6.22 mm

c)

Since there is no acceleration along the x-direction

Vx = final horizontal velocity = v = 4.80 x 105 m/s

Along Y-direction , using the equation

vy = vo + a t

vy = 0 + (9.4 x 1011) (1.15 x 10-7)

vy = 1.1 x 105 m/s

so vf = (4.80 x 105) i + (1.1 x 105) j

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