A proton moves at 4.60 105 m/s in the horizontal direction. It enters a uniform

Question

A proton moves at 4.60 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 7.60 103 N/C. Ignore any gravitational effects.

(a) Find the time interval required for the proton to travel 5.50 cm horizontally.
(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.)
(c) Find the horizontal and vertical components of its velocity after it has traveled 5.50 cm horizontally.

Explanation / Answer

A. The E-field has no effect on the proton's horizontal motion. So the time it takes to go 0.055 meters is: time = distance/velocity time = 0.055/(4.6*10^5) time = 1.42857*10^-7 sec B. The proton accelerates vertically because of the E-field: F=qE ma=qE a=(q/m)E a=95788341 *7.6*10^3 a=7.28 *10^11 m/s2 We have the vertical acceleration and we found the time in part A, now we need the displacement: y=(1/2)at^2 no initial y-velocity y=(1/2)(7.28 *10^11)(1.42857*10^-7)^2 y=7.429*10^-3 m C. The vertical velocity we can easily get: v=at since intial vertical velocity was zero v=(7.28 *10^11)(1.42857*10^-7) v=103 999.896 m/s The horizontal velocity was never changed v=4.6*10^5 So we have (vx,vy)=(4.6*10^5, 1.04*10^5) m/s

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