A proton moves at 4.40 105 m/s in the horizontal direction. It enters a uniform

Question

A proton moves at 4.40 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.00 103 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 = < i + j > km/s

Explanation / Answer

A. The E-field has no effect on the proton's horizontal motion. So the time it takes to go 0.06 meters is: time = distance/velocity time = 0.06/(4.2*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=4.2*10^5 So we have (vx,vy)=(4.2*10^5, 1.04*10^5)

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