Question

In the figure, a proton is projected horizontally midway between two parallel plates that are separated by 0.50 cm, and are 5.60 cm long. The electrical field due to the plates has magnitude 610,000 N/C between the plates away from the edges (e = 1.60 x 10-19 C, ao = 8.85 x 10-12 C2/N-m2, mp = 1.67 x 10-27 kg) 111LE (a) Find the minimum speed of the proton if it just misses the lower plate as it emerges from th field. (b) Find the angle of the velocity with respect to the horizon at the point where the proton j misses the lower plate and exits the field.

Answer 1

a) Minimum speed of the proton: $6.05\cdot 10^6 m/s$

b) Angle of the velocity: $\theta=-5.1^{\circ}$

Explanation:

a)

The proton experiences a vertical force due to the electric field, given by:

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the proton charge

$E=610,000 N/C$ is the magnitude of the electric field

The vertical acceleration of the proton is therefore

$a=\frac{qE}{m}$

where

$m=1.67\cdot 10^{-27}kg$ is its mass

Therefore, the vertical position of the proton at time t is

$y(t)=\frac{1}{2}at^2=\frac{1}{2}\frac{qE}{m}t^2$

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

$x(t)=v_0 t$

where $v_0$ is the initial speed. This equation can be rewritten as

$t=\frac{x(t)}{v_0}$

And substituting into the eq. for y,

$y(t) = \frac{1}{2}\frac{qE}{m} \frac{x^2}{v_0^2}$

Solving for the initial speed,

$v_0 = \sqrt{\frac{qEx^2}{2my}}$

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:

$v=\sqrt{\frac{(1.6\cdot 10^{-19})(610,000)(0.056)^2}{2(1.67\cdot 10^{-27})(0.0025)}}=6.05\cdot 10^6 m/s$

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:

$v_x = v_0 = 6.05\cdot 10^6 m/s$

The vertical velocity is given by:

$v_y^2 - u_y^2 = 2ay$

where:

$u_y=0$ (initial vertical velocity is zero)

$a=\frac{qE}{m}$ (acceleration)

y = 0.0025 m (vertical displacement)

Solving for $v_y$,

$v_y = \sqrt{2ay}=\sqrt{2\frac{qEy}{m}}=5.4\cdot 10^5 m/s$

Therefore, the final angle of the velocity with respect to the horizontal is:

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.4\cdot 10^5}{6.05\cdot 10^6})=5.1^{\circ}$

And since the electric field is downward (the proton just misses the lower field), it means that the angle is below the horizontal:

$\theta=-5.1^{\circ}$

Learn more about electric fields:

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