A. A proton has an initial velocity of 1.64 × 10^7 m/s in thehorizontal directio
ID: 1755814 • Letter: A
Question
A. A proton has an initial velocity of 1.64 × 10^7 m/s in thehorizontal direction. It enters a uniform electric ?eld of 17200N/C directed vertically. Ignoring gravitational e?ects, ?nd thetime it takes the proton to travel 0.02 m horizontally. Answer inunits of ns.B. What is the vertical displacement of the proton after theelectric ?eld acts on it for that time? Answer in units of mm.
C. What is the proton’s speed after being in the electric?eld for that time? Answer in units of km/s.
I figured out part A which is 1.2195ns, or 1.2195e-9s & B(0.001223mm), but am stuck on C. Could somebody help me out please!
Explanation / Answer
initial velocity v = 1.64 × 10^7m/suniform electric feld E = 17200 N/C Distance S = 0.02 m Accleration in vertical direction a = Eq / m where q = charge = 1.6 * 10 ^ -19 C m =mass = 1.67 * 10 ^ -27 kg substitue values weget a = 1.6479 * 10 ^ 12 m / s^ 2 time taken to travel S distance is t =S / v Since there is no accleration alonghorizontal direction = 1.219 * 10 ^ -9 s = 1.219 ns (b). Vertical direction : ------------------ Initial velocity u = 0 m / s Accleration a = 1.6479 * 10 ^ 12 m / s ^ 2 time t= 1.219* 10 ^ -9 s from the relation y = ut + ( 1/ 2) at^ 2 vertical displacement y = 1.2243 * 10^ -6 m = 1.2243 * 10 ^ -3 mm =0.0012243 mm (c). Horizontal component of velcoity after time t is V = 1.64 * 10 ^ 7 m / s = 1.64 * 10 ^ 4 km / s Since there is no accleration Vertical component of velocity after time t is V ' = u+ at = 0 + 2 * 10 ^ 3 = 2 km / s Speed after time t is V " = [ V ^ 2 + V ' ^2 ] = 16400.000123 km /s (c).
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