Question

A proton with the mass m is projected into a uniform electric field that points vertically upward and has magnitude E. The initial velocity of the proton has a magnitude v0 and is directed at an angle ? (alpha) below the horizontal.A)Find the maximum distance hmax that the proton descends vertically below its initial elevation. You can ignore gravitational forces.b)Express your answer in terms of the variables m, ?, v0, E, e.After what horizontal distance d does the proton return to its original elevation?Express your answer in terms of the variables m, ?, v0, E, e.C)Find the numerical value of hmax if E = 400N/C , v0 = 5.00

Answer 1

Answer:

maximum distance h max  = Vo² (sin α)²   m/ 2 qE

Explanation:

We must know the acceleration that the electric field creates on the proton, we use Newton's second law, with the electric force  ( F= qE)

F = ma

qE = ma

a = qE / m

We use the kinematic equations for the y axis

Vy² = Voy² - 2 a Y

At the point of maximum descent Vy is zero

h max =Y=  Voy² /2 a

h max  = Vo² (sin α)²   m/ 2 qE

This is the point of maximum descent of the proton since the speed is zero and from here it starts to rise

b) For this point it is equivalent to the calculation of the range in the movement of projectiles with the given acceleration

R = Vo² sin (2α) / a

R = Vo² sin (2α) m / qE

c) We find the numerical value for the given values

    h max = 5² (sin α)² 1.67 10⁻²⁷ / 2 (1.6 10⁻19 400)

Answer 2

Answer:

distance = $\frac{m_{p}V_{o}^{2} sin2\alpha }{eE}$

Explanation:

the distance is given as:

acceleration:

$\frac{d}{2} = \frac{V_{o}cos\alpha }{V_{ox} }$

Thus, t is given by:

$t = \frac{m_{p}V^{2}sin\alpha cos\alpha }{eE}$

Thus,

$d= \frac{2M_{p}V_{o}^{2}sin\alphacos\alpha }{eE}$

  = $\frac{m_{p}V_{o}^{2} sin 2\alpha }{eE}$