A proton moves at 4.40 X 10^5 m/s in the horizontal direction. It enters a unifo

Question

A proton moves at 4.40 X 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 7.60 X 10^3 N/C. Ignore any gravitational effects.

(a) Find the time interval required for the proton to travel 6.00 cm horizontally.

______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.)

______mm

(c) Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally.

v with arrow = ____i + _____ j km/s

Explanation / Answer

a.)

as electric field is perpendicular to the velocity of charge so force exert in horizontal direction by electric field become zero

time interval required for the proton to travel 6.00 cm horizontally = Distance / speed

= 0.06 m / (4.40 X 10^5 m/s ) = 1.3636 *10^-7 s = 136.36 ns

b.)

force exert in verical direction = qE = 1.6 * 10^ -19 C * 7.60 X 10^3 N/C = 1.216 * 10^-15 N

using equn of motion

F = ma

a=  1.216 * 10^-15 N / 9.1 * 10^-27

a= 1.336 *10^11 m/s

s = ut + 0.5 a*t^2

s = 0 + 0.5 * 1.336 *10^11 m/s *(1.3636 *10^-7)^2 s = 1.242 * 10^-3 m = 1.242 mm

c.)

using equnation of motion

vertical velocity(Vy) = u + at

Vy = 0 + 1.336 *10^11 * 1.3636 *10^-7 s = 1.859 * 10^4 m/s

Vx = 4.40 X 10^5 m/s

v with arrow =  4.40 X 10^2 i +  1.859 * 10^1 j Km/s

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