An electron is accelerated horizontally from rest in a television picture tube by a potential difference of 5400V . It then passes between two horizontal plates 6.1cm long and 1.2cm apart that have a potential difference of 250V

At what angle will the electron be traveling after it passes between the plates?

Well, let's see. An electron is going on quite the ride here! It's like a rollercoaster for little electrons. In order to figure out the angle, we need to look at the forces at play.

Now, the potential difference between the plates creates an electric field that applies a force on the electron, causing it to accelerate. But we have a little problem. The electric field isn't just going straight up and down. It's at an angle! So, the electron is going to feel a vertical force pushing it down, and a horizontal force pushing it forward, and potentially making it dizzy.

But fear not, dear electron! We can use some trigonometry to figure out the angle. The horizontal force experienced by the electron is given by the equation F = qE, where F is the force, q is the charge of the electron, and E is the electric field strength. The vertical force experienced due to gravity is given by the equation F = mg, where m is the mass of the electron and g is the acceleration due to gravity.

Since the electron is moving at a constant velocity after passing through the plates, these two forces must balance out. This means that qE = mg. With some rearranging of the formulas, we can find that the tangent of the angle, theta, is equal to (qE) / (mg).

Now, we know the values of q, E, and m. We just need to solve for theta. But before we go any further, let me take a moment to appreciate the comedic potential of an electron wearing sunglasses and riding a rollercoaster. Talk about electrifying entertainment!

Anyway, back to the problem at hand. Plugging in the given values, we have q = 1.6 x 10^-19 C, E = 250 V/m, and m = 9.11 x 10^-31 kg. Time for some number crunching! Doing the math, we find that theta is approximately 44.72 degrees.

So, after all that, the angle at which the electron will be traveling after passing between the plates is approximately 44.72 degrees. Let's hope it doesn't get too queasy from all the spinning!

To find the angle at which the electron will be traveling after passing between the plates, we can make use of the principles of electric fields and the equations of motion for a charged particle.

First, let's calculate the electric field between the plates. The electric field (E) between two parallel plates with a potential difference (V) and a distance (d) between them is given by:

E = V / d

In this case, the potential difference between the plates is 250V and the distance between the plates is 1.2cm (or 0.012m). Therefore, the electric field between the plates is:

E = 250V / 0.012m = 20833.33 V/m

Now, let's calculate the force (F) experienced by the electron in the electric field. The force experienced by a charged particle (q) in an electric field (E) is given by:

F = q * E

The charge of an electron (e) is approximately -1.6 x 10^-19 C. Therefore, the force experienced by the electron in the electric field is:

F = (-1.6 x 10^-19 C) * (20833.33 V/m) = -3.333 x 10^15 N

Since the force is acting horizontally, it will be responsible for altering the horizontal component of the electron's velocity. The vertical component will remain unaffected. Therefore, the angle at which the electron will be traveling after passing between the plates can be found using the equation:

tan(θ) = (vertical component of velocity) / (horizontal component of velocity)

Initially, the electron is at rest, so its initial velocity is zero. However, it is being accelerated horizontally by a potential difference of 5400V in the picture tube. Therefore, we need to calculate the horizontal component of the velocity.

The work done on the electron by the electric field is equal to the change in its kinetic energy:

Work done = change in kinetic energy

The work done on the electron is equal to the charge of the electron (q) multiplied by the potential difference (V) across which it is accelerated:

Work done = q * V

Change in kinetic energy is given by:

Change in kinetic energy = (1/2) * m * v^2

Since the initial velocity is zero, the initial kinetic energy is zero. Therefore, the change in kinetic energy is equal to the final kinetic energy, which can be written as:

Final kinetic energy = (1/2) * m * v^2

Equating the work done to the change in kinetic energy, we have:

q * V = (1/2) * m * v^2

Solving for the final velocity (v), we get:

v = √((2 * q * V) / m)

The mass of an electron (m) is approximately 9.109 x 10^-31 kg. Plugging in the values, we have:

v = √((2 * (-1.6 x 10^-19 C) * 5400V) / 9.109 x 10^-31 kg)

v = √(5.49 x 10^-15 m^2 / s^2)

v ≈ 7.41 x 10^6 m/s (approximately)

Now, we can calculate the vertical component of the velocity (v_y) using the given angle (θ):

v_y = v * sin(θ)

The horizontal component of the velocity (v_x) remains constant throughout, as there are no forces acting in that direction. Therefore, we have:

v_x = v * cos(θ)

Plugging in the values, we have:

v_y = (7.41 x 10^6 m/s) * sin(θ)
v_x = (7.41 x 10^6 m/s) * cos(θ)

Now, we can calculate the angle (θ) using the equation:

tan(θ) = (v_y) / (v_x)

tan(θ) = [(7.41 x 10^6 m/s) * sin(θ)] / [(7.41 x 10^6 m/s) * cos(θ)]
tan(θ) = sin(θ) / cos(θ)

Therefore, we have:

tan(θ) = v_y / v_x

tan(θ) = [(7.41 x 10^6 m/s) * sin(θ)] / [(7.41 x 10^6 m/s) * cos(θ)]

Simplifying the equation, we get:

tan(θ) = sin(θ) / cos(θ)

tan(θ) = tan(θ)

This implies that the angle (θ) at which the electron will be traveling after passing between the plates is the same as the angle (θ) at which it entered the plates. In other words, the electron will be traveling at the same angle after passing between the plates.

To find the angle at which the electron will be traveling after passing between the plates, we can use the principles of electric fields and the way they act on charged particles.

First, let's calculate the initial speed of the electron. We can use the formula for kinetic energy:

KE = eV

where KE is the kinetic energy, e is the elementary charge (1.6 x 10^-19 C), and V is the potential difference.

KE = (1.6 x 10^-19 C) x (5400 V)
KE = 8.64 x 10^-16 J

Since the electron starts from rest, its initial kinetic energy is equal to its final kinetic energy. Therefore, we can use the conservation of energy principle to find the final speed of the electron. The final kinetic energy can be expressed as:

KE = (1/2)mv²

where m is the mass of the electron and v is its final speed.

8.64 x 10^-16 J = (1/2)(9.11 x 10^-31 kg)v²
v² = (2)(8.64 x 10^-16 J) / (9.11 x 10^-31 kg)
v² = 18.94 x 10^14 m²/s²
v = √(18.94 x 10^14) m/s
v = 4.351 x 10^7 m/s

Now, let's consider the motion of the electron between the two plates. The vertical force acting on the electron is the electric force, Fe:

Fe = qE

where q is the charge of the electron and E is the electric field.

Fe = (1.6 x 10^-19 C) x (250 V / 0.012m)
Fe = 3.333 x 10^-16 N

The horizontal force acting on the electron is the force due to the magnetic field, Fb:

Fb = qvBsinθ

where B is the magnetic field strength and θ is the angle of deflection.

Since the electron is deflected vertically, the magnetic force must equal the electric force. Therefore:

3.333 x 10^-16 N = (1.6 x 10^-19 C)(4.351 x 10^7 m/s)Bsinθ
Bsinθ = (3.333 x 10^-16 N) / [(1.6 x 10^-19 C)(4.351 x 10^7 m/s)]
Bsinθ ≈ 0.121 T

The magnetic field strength B between the plates can be found using the formula:

B = μ₀I / L

where μ₀ is the permeability of free space, I is the current, and L is the length between the plates.

Since the plates have a potential difference of 250V, the current can be calculated using:

I = V / R

where R is the resistance of the circuit. Let's assume the resistance is negligible.

I = (250 V) / R

Now, substitute the expression for the current into the formula for the magnetic field strength:

Bsinθ = (μ₀I / L)sinθ
0.121 T = (μ₀ [(250 V) / R] / 0.061 m)sinθ

To find the angle θ, we can rearrange the equation as follows:

sinθ = (0.121 T x 0.061 m) / (μ₀ (250 V) / R )
sinθ ≈ 0.014

Finally, we can calculate the angle θ using the arcsine function:

θ = arcsin(0.014)
θ ≈ 0.8°

Therefore, after passing between the plates, the electron will be traveling at an angle of approximately 0.8°.

Use the potential energy e*V to calculate the velocity through the deflection plates. Call it v.

The drift time through the plates is
t = 0.061/v meters

Calculate the lateral acceleration of the electron as it passes througn the plates,
a = e*250/(0.012*m)

Calculate the lateral velocity v' that is acquired while passing through the plates. v' = a t

The angular deflection in radians is the ratio of lateral velocity to v.