A particle of charge q, mass m, and zero initial speed, is accelerated by a pote

Question

A particle of charge q, mass m, and zero initial speed, is accelerated by a potential difference V_0 between two parallel flat grids spaced by a distance L = 5 cm. The potential is supposed varying in a linear manner along L. Give the velocity v of the particle when it will cross the second grid. What are the signs of both q and V_0 to make the particle being accelerated? As the potential is supposed varying in a linear manner along L. Find the time x of the path between the two grids. Using results found in (a) and (c), what are the values of v and tau for: an electron accelerated by V_0 = 100 V a proton accelerated by V_0 = -3000 V

Explanation / Answer

a. Given
charge of particle = q
mass = m
inital speed = 0 = initial KE
Potential difference = Vo
distance, L = 0.05 m
linear variation of V, so constant E

a. KE at second grid = KE at first grid + qV = qV
0.5mv^2 = qV
v = sqroot(2qV/m)
b. if q is +ve, Vo should be -ve
if q is -ve , Vo should be +ve
c. Potential is varying linearly, so the particle travelled with constant acceleration
2*a*L = v^2
2*a*L = 2qV/m
aL = qV/m
a = qV/mL [ a is aceleration of the particle]
v = at
sqroot(2qV/m) = qVt/mL
t = sqroot(2qV/m)/(qV/mL) = sqroot(2m/qV)/L
d. q = 1.6*10^-19 C
m = 9.1*10^-31 kg
Vo = 100 V
v = 5.929*10^6 m/s
t = 6.745*10^-6 s

m = 1.6*10^-27 kg
V = -3000 V
v = 0.7745*10^6 m/s
t = 5.163*10^-5 s

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