12. 4/8 points | Previous Answers SerPSE9 23.P.057.MI My Notes Ask Your Teacher

Question

12. 4/8 points | Previous Answers SerPSE9 23.P.057.MI My Notes Ask Your Teacher A proton moves at 4.20 x 10 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.20 x 103 N/C Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 6.00 cm horizontally. 143 ns (b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.) 8.99 (c) Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally ) km/s Need Help?

Explanation / Answer

vix=4.20*10^5m/s, E=9.20*10^3 N/C, X= 0.06m

Vertical force due to electric field E = F= qE = (1.6*10^-19)*(9.20*10^3) = 1.47*10^-15 N

Acceleration in the direction of force = a= F/m = (1.47*10^-15)/(1.67*10^-27) = 8.80*10^11 m/s^2

a)

X=vix*t

t= X/vix= (0.06)/(4.20*10^5) = 143*10^-9 s

b)

y=viy*t +1/2*at^2 = 0*1.43*10^-4 +1/2*(8.80*10^11)*(1.43*10^-9)^2 = 8.997m m

c) Since there is no acceleration in horizontal direction vfx=vix= 4.20*10^5 m/s

Along y direction,

vfy=viy+at = 0+(8.80*10^11)*(1.43*10^-9) = 1.258*10^3 = 1.26km/s

Thus v=(420.0)i+(1.26)j   km/s

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