A proton moves at 4.20 * 10 5 m/s in the horizontal direction. It enters a unifo

Question

A proton moves at 4.20 * 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 * 103 N/C. Ignore any gravitational effects.

(a) Find the time interval required for the proton to travel 5.00 cm horizontally.

(b) Find its vertical displacement during the time interval in which it travels 5.00 cm horizontally.

(c) Find the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.

If anyone helps me, I will reward points.

Explanation / Answer

the accelaration along horizontal direction is zero,since the force on the proton is only in vertical direction,

u=4.20 * 105 m/s
(a) time taken = 5*10^-2/u = 5*10^-2/4.20 * 105 = 1.19*10-7 s = 119 ns

(b) electric field force on proton = 8.4*103*1.6*10-19 N

  from s=ut+1/2at2

s = 0.5 a* (119*10-9)2

accelaration in vertical direction a = F/m = 8.4*103*1.6*10-19/(1.672*10-27) = 8.04*1011 m/s2

s= 0.5*8.04*1011*(119*10-9)2 = 5.69*10-3 m = 0.569 cm

(c) vhorizontal = 4.20 * 105 m/s

vvertical = 0 + at = 8.04*1011*119*10-9

= 95676 m/s

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