91Ó°ÊÓ

Given is E = 500 N/C, v= 4.0 x 10³ m/, m = 1.67 × 107 kg,α = 30°

a) As previously mentioned the motion of the charge in a parallel plate electric field is similar to projectile motion

The initial velocity in the d-direction is given by

${\mathrm{v}}_{\mathrm{x}}={\mathrm{v}}_0\cos (\mathrm{Î\pm })=\frac{\mathrm{d}}{\mathrm{t}}⇒{\mathrm{v}}_0=\frac{\mathrm{d}}{\mathrm{t}\mathrm{³¦´Ç²õÎ\pm }}$ (1)

The proton will move$h_{max}$in half of the total time and its velocity will be instantaneously zero then it will begin moving in a vertical direction until reaching the original takeoff point with the same initial velocity component so this displacement $h_{max}$is given by

$h_{max}=\frac{{v_f}^2}{2a_{v_{max}}}$ (2)

in the upward direction, the initial velocity is zero, the final velocity is the same as the initial component and the acceleration is given by

$F_{net}=m{a}_v=eE⇒a_v=\frac{eE}{m}$ (3)

So substitution in (2) yields

${\mathrm{h}}_{\max }={\frac{\mathrm{m}({\mathrm{v}}_0\mathrm{²õ¾\pm ²ÔÎ\pm })}{2\mathrm{eE}}}^2=\frac{{{\mathrm{mv}}^2}_0{\sin }^2\mathrm{Î\pm }}{2\mathrm{eE}}$………. (4)

b) To get the displacement d traveled in terms of constant first we must know that

${\mathrm{h}}_{\max }={\mathrm{v}}_{\mathrm{by}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\left(\frac{\mathrm{t}}{2}\right)}^2$ (5)

Substitution from (1) and (4) in (5) yields

${{\mathrm{mv}}^2}_0{\sin }^2\mathrm{Î\pm }=\frac{1}{2}×\frac{\mathrm{eE}}{\mathrm{m}}×\left(\frac{\mathrm{d}}{2{\mathrm{v}}_0\mathrm{³¦´Ç²õÎ\pm }}\right)$

The displacement is given by

b) To get the displacement d traveled in terms of constant first we must know that

${\mathrm{h}}_{\max }={\mathrm{v}}_{\mathrm{ay}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\left(\frac{\mathrm{t}}{2}\right)}^2$

Substitution from (1) and (4) in (5) yields

${{\mathrm{mv}}^2}_0{\sin }^2\mathrm{Î\pm }=\frac{1}{2}×\frac{\mathrm{eE}}{\mathrm{m}}×\left(\frac{\mathrm{d}}{2{\mathrm{v}}_0\mathrm{³¦´Ç²õÎ\pm }}\right)$

The displacement is given by

$\frac{\mathrm{d}}{2{\mathrm{v}}_0\mathrm{³¦´Ç²õÎ\pm }}=\frac{{{\mathrm{mv}}^2}_0\mathrm{²õ¾\pm ²ÔÎ\pm }}{2\mathrm{eE}}⇒\mathrm{d}=\frac{{{\mathrm{mv}}^2}_0\sin \left(2\mathrm{Î\pm }\right)}{\mathrm{eE}}$ (6)

Taking into account that: $2\mathrm{²õ¾\pm ²ÔÎ\pm }\mathrm{³¦´Ç²õÎ\pm }=\sin 2\mathrm{Î\pm }$

c) In equation (2) if we substitute from (1) we get that

The previous trajectory is an equation of parabola so the trajectory

d) The numerical value of ${\mathrm{h}}_{\max }$is given by

Also, the numerical value of d is

${\mathrm{h}}_{\max }=\frac{\left(1.67×{10}^{-27}\right)×{\left(4×10\right)}^5×\left(\sin 60\right)}{1.6×{10}^{-14}×500}=2.89$