Respuesta :
a) Minimum speed of the proton: [tex]6.05\cdot 10^6 m/s[/tex]
b) Angle of the velocity: [tex]\theta=-5.1^{\circ}[/tex]
Explanation:
a)
The proton experiences a vertical force due to the electric field, given by:
[tex]F=qE[/tex]
where
[tex]q=1.6\cdot 10^{-19}C[/tex] is the proton charge
[tex]E=610,000 N/C[/tex] is the magnitude of the electric field
The vertical acceleration of the proton is therefore
[tex]a=\frac{qE}{m}[/tex]
where
[tex]m=1.67\cdot 10^{-27}kg[/tex] is its mass
Therefore, the vertical position of the proton at time t is
[tex]y(t)=\frac{1}{2}at^2=\frac{1}{2}\frac{qE}{m}t^2[/tex]
where we assumed that the initial vertical velocity is zero (because the proton is fired horizontally) and the initial vertical position, halfway between the two plates, is the origin.
The horizontal motion of the proton instead is uniform, so the horizontal position is given by
[tex]x(t)=v_0 t[/tex]
where [tex]v_0[/tex] is the initial speed. This equation can be rewritten as
[tex]t=\frac{x(t)}{v_0}[/tex]
And substituting into the eq. for y,
[tex]y(t) = \frac{1}{2}\frac{qE}{m} \frac{x^2}{v_0^2}[/tex]
Solving for the initial speed,
[tex]v_0 = \sqrt{\frac{qEx^2}{2my}}[/tex]
The proton just misses one of the plate when
x = 5.60 cm = 0.056 m (length of the plates)
y = 0.25 cm = 0.0025 m (half the distance between the plates)
Therefore, we find the initial speed:
[tex]v=\sqrt{\frac{(1.6\cdot 10^{-19})(610,000)(0.056)^2}{2(1.67\cdot 10^{-27})(0.0025)}}=6.05\cdot 10^6 m/s[/tex]
b)
In order to find the angle, we just need to analyze the horizontal and vertical component of the final velocity of the proton.
The horizontal velocity is constant so it is:
[tex]v_x = v_0 = 6.05\cdot 10^6 m/s[/tex]
The vertical velocity is given by:
[tex]v_y^2 - u_y^2 = 2ay[/tex]
where:
[tex]u_y=0[/tex] (initial vertical velocity is zero)
[tex]a=\frac{qE}{m}[/tex] (acceleration)
y = 0.0025 m (vertical displacement)
Solving for [tex]v_y[/tex],
[tex]v_y = \sqrt{2ay}=\sqrt{2\frac{qEy}{m}}=5.4\cdot 10^5 m/s[/tex]
Therefore, the final angle of the velocity with respect to the horizontal is:
[tex]\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.4\cdot 10^5}{6.05\cdot 10^6})=5.1^{\circ}[/tex]
And since the electric field is downward (the proton just misses the lower field), it means that the angle is below the horizontal:
[tex]\theta=-5.1^{\circ}[/tex]
Learn more about electric fields:
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The velocity depends on position and time interval.
Part A: The minimum speed of proton if it just misses the lower plate in the field is [tex]6.05\times 10^6\;\rm m/s[/tex].
Part B: The angle of velocity in the horizontal direction below the plate is -5.10 degrees.
What is velocity?
Velocity is defined as the change in position of an object with respect to a given time interval.
Given that the electric field E is 610,000 N/C and the charge q at the proton is 1.60 x 10-19 C.
Part A:
The vertical force acted on the proton due to the electric field is given as below.
[tex]F = qE[/tex]
We know that the force is given as,
[tex]F = ma[/tex]
Where a is the vertical acceleration and m is the mass of the proton. By equating both equations, we get,
[tex]ma=qE[/tex]
[tex]a = \dfrac {qE}{m}[/tex]
[tex]a = \dfrac { 1.60\times 10^{-19}\times 610000}{1.67\times 10^{-27}}[/tex]
[tex]a = 5.84\times 10^{13}\;\rm m/s^2[/tex]
The vertical position of the proton at time t is given as below.
[tex]s(t) = ut + \dfrac{1}{2}at^2[/tex]
Where u is the initial velocity of the proton. The proton is projected horizontally so its vertical initial position between the plates is its origin, hence its initial velocity will be zero.
[tex]s(t) = \dfrac{1}{2}at^2[/tex]
[tex]s(t) = \dfrac {1}{2}\times 5.84\times 10^{13} t^2[/tex]
[tex]s(t)=2.92\times 10^{13} t^2[/tex]
The horizontal position of the proton is given below.
[tex]s'(t) = u't + 0[/tex]
[tex]s'(t) = u't[/tex]
[tex]t = \dfrac{s'(t)}{u'}[/tex]
Where u' is initial velocity in the horizontal direction.
Substituting the value of t, we get,
[tex]s(t) = 2.92\times 10^{13} (\dfrac {s'(t)}{u'})^2[/tex]
[tex]s(t) = 2.92\times 10^{13} \times \dfrac { s'(t)^2}{u'^2}[/tex]
[tex]u' = \sqrt{2.92\times 10^{13} \times \dfrac {s'(t)^2 }{s(t)}}[/tex]
For horizontal direction, the length of the plate is 5.60 cm and for the vertical direction, the midpoint between the plates is 0.25 cm.
[tex]u' = \sqrt{2.92\times 10^{13} \times \dfrac {0.056^2}{0.0025}}[/tex]
[tex]u' = 6.05\times 10^6 \;\rm m/s[/tex]
Hence we can conclude that the minimum speed of proton if it just misses the lower plate in the field is [tex]6.05\times 10^6\;\rm m/s[/tex].
Part B:
The vertical velocity of the proton is calculated as given below.
[tex]u^2 - u'^2 = 2as(t)[/tex]
[tex]u^2 - 6.05\times 10^6 = 2\times 5.84\times 10^{13} \times 0.0025[/tex]
[tex]u^2 = 2.92 \times 10^{11}[/tex]
[tex]u = 5.4 \times 10^5 \;\rm m/s[/tex]
The final angle of velocity from the horizontal plane is calculated as given below.
[tex]\theta = tan^{-1} \dfrac {u}{u'}[/tex]
[tex]\theta = tan^{-1} \dfrac {5.4\times 10^5}{6.05\times 10^6}[/tex]
[tex]\theta = 5.10 ^\circ[/tex]
The proton just misses the lower field so the electric field is downward, it means that the angle is below the horizontal plane.
The angle of the velocity is -5.10 degrees.
To know more about the velocity, follow the link given below.
https://brainly.com/question/862972.
