Question

a uniform, upward-pointing electric field E of magnitude 4.00×103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 5 cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle θ=45° with the lower plate and has a magnitude of 7.86×106 m/s. Will the electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates ( m).

Answer 1

Using projectile motion and the concept of an electric field, it is calculated that the electron will strike the lower plate at a horizontal distance of approximately 0.050 m from the left edge.

Step-by-step explanation:

To calculate whether the electron will strike one of the plates, we need to use the concepts of electric fields and projectile motion. The problem involves a uniform electric field directed upward, and an electron with an initial velocity at a 45° angle relative to the horizontal (lower plate).

First, we need to decompose the initial velocity of the electron into its horizontal (v_x) and vertical (v_y) components:

• v_x = v_0 cos(θ) = (7.86 × 10^6 m/s) cos(45°) = 5.56 × 10^6 m/s
• v_y = v_0 sin(θ) = (7.86 × 10^6 m/s) sin(45°) = 5.56 × 10^6 m/s

Next, calculate the time taken by the electron to travel the length of the plates horizontally (time of flight t). Since the horizontal motion is not affected by the electric field:

t = L / v_x = 0.05 m / (5.56 × 10^6 m/s) ≈ 9.00 × 10^-9 s

Now, we apply Newton's second law to find the vertical acceleration due to the electric field. The force on the electron is F = eE, where e is the charge of the electron (1.60 × 10^-19 C) and E is the electric field magnitude (4.00 × 10^3 N/C).

a = F/m = eE/m_e = (1.60 × 10^-19 C)(4.00 × 10^3 N/C) / (9.11 × 10^-31 kg) ≈ 7.00 × 10^14 m/s^2

Because the force is upward and the electron's charge is negative, the acceleration is downward:

a_y = -a = -7.00 × 10^14 m/s^2

Now we can determine the vertical displacement (y) using the kinematic equation:

y = v_yt + (1/2)a_yt^2

y = (5.56 × 10^6 m/s)(9.00 × 10^-9 s) + (1/2)(-7.00 × 10^14 m/s^2)(9.00 × 10^-9 s)^2

y ≈ -0.0285 m

Since the separation between the plates is 0.02 m and the vertical displacement y is larger in magnitude than the separation, the electron will indeed strike the lower plate. To find the horizontal distance from the left edge, we use the flight time, t:

x = v_xt = (5.56 × 10^6 m/s)(9.00 × 10^-9 s) ≈ 0.050 m

The electron will therefore strike the lower plate at a horizontal distance of approximately 0.050 m from the left edge.

Answer 2

As the distance is greater than the separation of the electron plates collides with it

X1= 3.02 10⁻² m

Step-by-step explanation:

To solve that problem we will work with Newton's second law to find acceleration and then we will use the kinematic equations

Electric force is

F = E q

Where E is the electric field and q the charge in this case the charge of an electron e = 1.6 10-19 C, of negative sign and its mass is 3.1 10-31 Kg

F = ma

-E e= m a

a = -Ee / m

As the force is attractive it is directed on the Y axis

a = - Ee / m j ^

a = -4.00 10³ 1.6 10⁻¹⁹ / 3.1 10⁻³¹

a = -2.06 10¹⁵ m / s² j ^

Let's use the parabolic motion equations to find the position of the electron at the end of the plates, let's decompose the velocities

Vox = Vo cos θ

Voy = Vo sin θ

Vox = 7.86 10⁶ cos (-45)

Voy = 7.86 10⁶ sin (-45)

Vox = 5.56 10⁶ m / s

Voy = -5.56 10⁶ m / s

The movement in the X axis is uniform, so we can find the time it takes to cross the plate

X = Vox t

t = X / Vox

t = 5 10⁻² / 5.56 10⁶

t = 8.99 10⁻⁹ s

With this time we can see how much the electron descended

Y = Voy t - ½ a t²

Y = -5.56 10⁶ 8.99 10⁻⁹ - ½ 2.06 10¹⁵ (8.99 10⁻⁹)²

Y = -49.98 10⁻³ - 83.24 10⁻³

Y = -133.22 10⁻³ m = 1.33 10⁻¹ m

In the exercise they give as data that the separation of the plates is 2.00 cm, let's reduce the distance found to centimeters

Y = 133.22 10⁻³ m (100cm / 1m) = 13,322 cm

As the distance is greater than the separation of the electron plates collides with it

Second Part

Let's look for the time it takes to lower the width of the plate

Y = 2.00 cm = 2.00 10⁻² m

Y = Voy t - ½ a t²

-2.00 10⁻² = -5.56 10⁶ t - ½ 2.06 10¹⁵ t²

0 = 2 10⁻² + ​​5.56 10⁶ t - 1.03 10¹⁵ t²

1.03 10¹⁵ t² - 5.56 10⁶ t - 2 10⁻² = 0

We solve the second degree equation

t = 5.56 10⁶ √ [(5.56 10⁶) 2 - 4 1.03 10¹⁵ 2.10⁻²]} / (2 1.03 10¹⁵)

t = 5.56 10⁶ ±√ [30.91 10¹² - 8.24 10¹¹]} / 2.06 10¹⁵

t = 5.56 10⁶ ± 5.63 10⁶} /2.06 10¹⁵

t1 = 11.19 10⁶ /2.06 1015

t1 = 5.43 10⁻⁹ s

t2= -0.07 10⁻⁹ s

We take the right time as the positive. Let's calculate the horizontal distance

Xi = Vox t1

X1 = 5.56 10⁶ 5.43 10⁻⁹

X1 = 30.19 10⁻³ m

X1= 3.02 10⁻² m