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

The rate of change of velocity with respect to time is known as acceleration, The value  $\rm h_{max}$ will be $V_0^2 (sin \alpha)^2 \frac{m}{2} qE$.

What is acceleration?

The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

(a) The maximum distance that the proton descends vertically below its initial elevation is $V_o^2 (sin \alpha)^2 \times \frac{m}{2 qE}$.

The value of the acceleration of proton is found by ;

$\rm F = ma\\\\qE = ma\\\\ a = \frac{ qE}{m}$

As the value of the acceleration is constant, Newton's second equation of motion states at the top;

$\rm V_y^2 = V_oy^2- 2 a Y$

At the maximum height, the velocity is zero.

$h_{max }=Y= V_0y^2 /2 a\\\\ h_{max } = V_o^2 (sin \alpha)^2 \times \frac{m}{2 qE}$

Hence the value  $\rm h_{max}$ will be $V_0^2 (sin \alpha)^2 \frac{m}{2} qE$.

b) For this point it is equivalent to the calculation of the range in the movement of projectiles with the given acceleration is $V_o^2 sin (2\alpha) \frac{m}{qE}$.

The given equation is;

$R = V_o^2 \frac{sin (2]\alpha) }{a} \\\\ R = V_o^2 sin (2\alpha) \frac{m}{qE}$

Hence the range in the movement of projectiles with the given acceleration is $V_o^2 sin (2\alpha) \frac{m}{qE}$

(c) The numerical value  $h_{max}$ will be $3.29 \times 10^9 sin^2\alpha$.

The numerical value of hmax if E = 400N/C , v₀ = 5.00

$h_{max } = 5^2 (sin \alpha)^2 \times \frac{1.67 \times 10^{-27}}{2 \times 1.6 \times 10^{-19}\times 400} \\\\ h_{max } =3.29 \times 10^9 sin^2\alpha$

Hence the numerical value of maximum height will be $3.29 \times 10^9 sin^2\alpha$

To learn more about the acceleration refer to the link;

https://brainly.com/question/12134554