Until recently, TV sets consisted of what was called a cathode-ray tube. Electrons were boiled off a light bulb filament and passed through plates that deflected the particles; the electron beam went on to strike the TV screen itself and lit up a pixel of phosphor. The beam could be deflected both up and down and sideways (only one direction shown). By deflecting the electron beam and scanning it across the screen a picture was built up. Consider the simplified TV below. An electron beam is shot along the centerline between the two deflector plates. The top plate is held by a power supply (not shown) at a voltage \(V=\) \(+1000\) volts above the bottom plate. The separation between the plates is \(d=4 \mathrm{~cm}\). The length of the plates is \(\ell=8 \mathrm{~cm}\). The distance \(L\) from the end of the deflector plate to the TV screen is \(30 \mathrm{~cm}\). The electrons enter the region between the deflector plates with a velocity \(v=4 \times 10^7 \mathrm{~m} / \mathrm{s}\). a) In which direction is the electric field between the deflector plates? b) In which direction are the electrons deflected? c) What is the force on an electron? (Neglect gravity.) d) What is the acceleration on the electron? e) What is the total deflection of the electron as it hits the TV screen? f) What size magnetic field and in what direction would you need to place between the deflector plates in order to prevent the electrons from being deflected?

Short Answer

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Question: In a cathode-ray tube, an electron is moving horizontally with a velocity of 4 x 10^7 m/s between two deflector plates with a potential difference of 1000V and a distance of 0.04 m apart. The length of the plates is 0.08 m. Calculate the total deflection of the electron as it hits the TV screen and the size and direction of the magnetic field needed to prevent electron deflection. Answer: The total deflection of the electron as it hits the TV screen is 8.8 mm in the upward direction. To prevent electron deflection, a magnetic field of 6.25 x 10^-4 T directed into the plane (out of the page, following the right-hand rule) is needed.

Step by step solution

01

Analyzing the deflector plates

The top deflector plate has a positive voltage of +1000V, and the bottom plate is at a lower potential. Therefore, the electric field direction between the plates is from the top plate to the bottom plate. #b: Direction of electron deflection#
02

Analyzing the forces on the electron

The electron is negatively charged, and it moves in the presence of an electric field. As the electric field is directed from the top plate to the bottom plate, the force on the electron will be opposite to the electric field, i.e., from the bottom plate to the top plate. Thus, the electron will be deflected upwards. #c: Force on the electron#
03

Calculating the electric field and force on the electron

To calculate the force on the electron, we first need to find the electric field intensity between the plates. We know that electric field intensity E can be calculated as E = V/d, where V is the potential difference and d is the distance between the plates. Substituting the given values, E = 1000 V / 0.04 m = 25,000 V/m. Now we can calculate the force on the electron using F = qE, where q is the electron's charge. The charge of an electron (q) is approximately -1.6 x 10^-19 C. So, F = (-1.6 x 10^-19 C) * (25000 V/m) = -4 x 10^-15 N. Thus, the force on the electron is 4 x 10^-15 N in the upward direction (ignoring the negative sign as it just indicates direction). #d: Acceleration on the electron#
04

Calculating the acceleration of the electron

Using Newton's second law, F = ma, where m is the mass of the electron and a is its acceleration. The mass of an electron is approximately 9.1 x 10^-31 kg. We can rearrange this equation to solve for acceleration: a = F/m. Substituting the values, a = (4 x 10^-15 N) / (9.1 x 10^-31 kg) = 4.4 x 10^15 m/s^2. Hence, the acceleration in the upward direction is 4.4 x 10^15 m/s^2. #e: Total deflection of the electron#
05

Calculating the total deflection of the electron

The deflection can be calculated using kinematic equations. We know that the time spent by the electron inside the plates is t = l/v, where l is the length of the plates, and v is the horizontal velocity. Substituting the given values, t = 0.08 m / (4 x 10^7 m/s) = 2 x 10^-9 s. Now, using the equation for distance covered under constant acceleration, we have: Deflection (h) = 0.5 * a * t^2, where a is the upward acceleration and t is the time spent between the plates. h = 0.5 * (4.4 x 10^15 m/s^2) * (2 x 10^-9 s)^2 = 8.8 x 10^-3 m. Thus, the total deflection of the electron as it hits the TV screen is 8.8 mm in the upward direction. #f: Size and direction of magnetic field#
06

Calculating the magnetic field to prevent electron deflection

To prevent the electron deflection, we need to find a magnetic field that produces an equal but opposite magnetic force on the electron. The magnetic force can be calculated using the Lorentz force formula: F_m = qvB, where B is the magnetic field. To negate the effect of the electric field, we need F_m = F, or qvB = qE. Since q = e = -1.6 x 10^-19 C, we can write: B = E/v Substituting the values, B = (25,000 V/m) / (4 x 10^7 m/s) = 6.25 x 10^-4 T. Thus, a magnetic field of 6.25 x 10^-4 T directed into the plane (out of the page, following the right-hand rule) is needed to prevent the electrons from being deflected.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field Direction
Understanding the direction of the electric field is crucial in determining how charged particles like electrons will interact within that field. In the context of our cathode-ray tube scenario, the electric field is created between two deflector plates. Since the top plate has a higher potential (positive) than the bottom one, the electric field lines point from the top plate to the bottom plate. This implies that any positive charge would naturally move in the direction of the field, while a negative charge, such as an electron, would be pushed in the opposite direction. It's the fundamental behavior of electric charges: like charges repel, while opposite charges attract.

Thus, for our electron beam travelling between these plates, it enters a region where the electric field is oriented vertically, from positive to negative, influencing the path the electrons take as they move forward.
Electron Deflection
As for the actual movement of electrons, their deflection within the cathode-ray tube is a direct result of the electric field's influence. Given that the electric field direction is from the top plate to the bottom plate, and electrons are negatively charged, they experience a force in the opposite direction. This leads to their deflection upwards. This deflection allows a cathode-ray tube to control where the electron beam strikes, illuminating specific pixels on the screen.

The amount of deflection can be tuned by adjusting the voltage across the deflector plates, giving the user or device control over the electron beam's trajectory, crucial for creating the desired images on the screen.
Force on an Electron
Considering the impact of the electric field on an electron, we can quantify the force exerted on it using the expression: \( F = qE \), where \( q \) is the charge of the electron (-1.6 x 10^-19 C), and \( E \) is the electric field intensity, calculated by the potential difference over the distance between the plates. This interaction is fundamental to not just our television example but to many applications in electronics and particle physics where electric fields control the trajectory of charged particles.

By using the known values of the electric field and the charge of an electron, we can determine the exact magnitude of force that will act upon it in the given setup, allowing for precise predictions about its motion.
Electron Acceleration
Acceleration is a key concept when predicting how quickly an object changes its velocity. For an electron within an electric field, this acceleration follows Newton's second law \( F = ma \). Here, \( m \) is the mass of the electron and \( a \) is its acceleration. By rearranging the equation we find \( a = F/m \), which allows us to calculate how quickly the electron will change its velocity due to the force we've just calculated. In the case of the cathode-ray tube, this acceleration determines how fast the electron will move away from its initial path due to the electric field between the deflector plates.

The extreme values of acceleration are typical for particles at the microscopic level, such as electrons, and are a testament to the powerful forces at play at such scales.
Kinematic Equations in Electric Fields
The movement of electrons within an electric field can be beautifully described using kinematic equations, which traditionally detail the motion of objects under constant acceleration. In our cathode-ray tube example, once we've established the electron's acceleration, we can predict its subsequent path. To find the total deflection, we calculate the time the electron is under acceleration and then apply the kinematic equation \( h = 0.5 \times a \times t^2 \), where \( h \) is the deflection, \( a \) is the acceleration, and \( t \) is the time.

Even though these equations were first defined for large-scale objects, they apply just as well to subatomic particles like electrons, offering a deep insight into the deterministic nature of classical physics, spanning scales from falling apples to steering electrons.
Lorentz Force
The Lorentz force is the combined electric and magnetic force on a point charge due to electromagnetic fields. In the context of our electron beam, to counteract the effect of the electric field and prevent deflection, we would apply a magnetic field. Using the Lorentz force equation \( F_m = qvB \), where \( F_m \) is the magnetic force, \( q \) is the charge of the electron, \( v \) is the velocity of the electron, and \( B \) is the magnetic field strength, we can calculate the exact strength and orientation of a magnetic field needed. The right-hand rule helps us determine the direction of the magnetic field: for the electron's upward deflection to be negated, the magnetic field should be directed into the plane of the screen.

This aspect of the Lorentz force is widely used in technology and research to manipulate charged particle beams, including in cathode-ray tubes, particle accelerators, and even to confine plasma in fusion reactors.

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Most popular questions from this chapter

Two identical conducting spheres, \(A\) and \(B\) carry charges \(+Q\) and \(-Q\), respectively. A third, identical conducting sphere \(C\) carries charge \(Q=0\). Sphere \(A\) is touched to sphere \(C\) and separated. Next, sphere \(B\) is touched to sphere \(C\) and separated. Finally, \(A\) is touched to \(B\) and separated. What is the final charge on each sphere? (A) \(A=Q ; B=-Q ; C=0\) (B) \(A=Q / 2 ; B=Q / 2 ; C=Q / 4\) (C) \(A=Q / 8 ; B=Q / 8 ; C=-Q / 4\) (D) \(A=Q / 2 ; B=-Q / 4 ; C=-Q / 4\) (E) \(A=Q / 4 ; B=-Q / 8 ; C=-Q / 4\)

Two positive charges of magnitude \(q\) and \(2 q\) are fixed in place along the \(x\)-axis. Is there any place along the \(x\)-axis where the total field could be zero? (A) Yes, somewhere to the left of the charge \(q\) (B) Yes, somewhere to the right of the charge \(2 q\) (C) Yes, between the two charges but closer to \(q\) (D) Yes, between the two charges but closer to \(2 q\) (E) No, the field can never be zero

Proton \(A\) is moving with speed \(10^6 \mathrm{~m} / \mathrm{s}\) in a magnetic field of \(0.01 \mathrm{~T}\). Proton \(B\) is moving in the same magnetic field with speed \(2 \times 10^6 \mathrm{~m} / \mathrm{s}\). One can conclude that (A) Both protons have an orbital radius of about 1 meter. (B) Both protons have the same orbital period of about \(6.5 \times 10^{-6} \mathrm{~s}\). (C) proton \(B\) has twice the orbital period of proton \(A\). (D) proton \(B\) has twice the orbital frequency of proton \(A\). (E) proton \(B\) has half the orbital radius of proton \(B\).

Two identical plastic balls of mass \(10 \mathrm{gm}\) each are hung by threads with length \(30 \mathrm{~cm}\) from a common point, as shown below. The balls are each charged with the same charge \(q\) and repel each other until they come to rest with a horizontal separation of \(30 \mathrm{~cm}\). a) Sketch the electric field produced by the two balls. b) Draw the force vectors on the right-hand ball. c) What is the charge \(q\) in each ball?

Two long, parallel wires separated by a distance \(r\) carry equal currents \(I\) in opposite directions, as shown. The direction of the field caused by the top wire at the position of the bottom wire and the direction of the force exerted by the top wire on the bottom wire are (A) \(B\) into the page; \(F\) down (B) \(B\) up; \(F\) into the page (C) \(B\) into the page; \(F\) up (D) \(B\) out of the page; \(F\) down (E) \(B\) down; \(F\) out of the page

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