Question

An electron (mass = 9.11 \(\times\) 10\(^{-31}\) kg) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 cm away. It reaches the grid with a speed of 3.00 \(\times\) 10\(^6\) m/s. If the accelerating force is constant, compute (a) the acceleration; (b) the time to reach the grid; and (c) the net force, in newtons. Ignore the gravitational force on the electron.

Short answer

a) \(2.5 \times 10^{14}\) m/s², b) \(1.2 \times 10^{-8}\) s, c) \(2.28 \times 10^{-16}\) N.

Step-by-step solution

Identify the Known Variables

We know the initial speed of the electron, \(u = 0\) m/s, the final speed \(v = 3.00 \times 10^6\) m/s, the distance \(d = 1.80\) cm = 0.018 m, and the mass of the electron \(m = 9.11 \times 10^{-31}\) kg.

Apply the Equation for Acceleration

The equation to use is derived from the kinematic equation: \[ v^2 = u^2 + 2ad \]Rearrange the formula to solve for acceleration \(a\): \[ a = \frac{v^2 - u^2}{2d} \]Substitute the known values:\[ a = \frac{(3.00 \times 10^6)^2 - 0^2}{2 \times 0.018} \] Calculate \(a\).

Calculate the Acceleration

Using the formula from Step 2: \[ v^2 = (3.00 \times 10^6)^2 = 9.00 \times 10^{12} \] \[ a = \frac{9.00 \times 10^{12}}{2 \times 0.018} \] \[ a = \frac{9.00 \times 10^{12}}{0.036} \approx 2.5 \times 10^{14} \text{ m/s}^2 \]The acceleration of the electron is approximately \(2.5 \times 10^{14}\) m/s².

Calculate the Time to Reach the Grid

Use the formula relating velocity, acceleration, and time: \[ v = u + at \]Rearrange to solve for time \(t\): \[ t = \frac{v - u}{a} \]Substitute the known values:\[ t = \frac{3.00 \times 10^6 - 0}{2.5 \times 10^{14}} \] Calculate \(t\).

Solve for Time

Substitute and calculate:\[ t = \frac{3.00 \times 10^6}{2.5 \times 10^{14}} \approx 1.2 \times 10^{-8} \text{ seconds} \]The time to reach the grid is approximately \(1.2 \times 10^{-8}\) seconds.

Calculate the Net Force

Using Newton's second law,\[ F = ma \]Substitute the known values for mass \(m\) and acceleration \(a\):\[ F = 9.11 \times 10^{-31} \cdot 2.5 \times 10^{14} \]Calculate \(F\).

Solve for the Net Force

Substituting the values, we have:\[ F = 9.11 \times 10^{-31} \cdot 2.5 \times 10^{14} \approx 2.28 \times 10^{-16} \text{ N} \]The net force acting on the electron is approximately \(2.28 \times 10^{-16}\) newtons.

Key concepts

kinematic equations

Kinematic equations are essential tools for solving problems involving the motion of objects where acceleration is constant. These equations relate the initial and final velocities, acceleration, displacement, and time. One key kinematic equation used to find acceleration is:

• \[ v^2 = u^2 + 2ad \]

Here, \(v\) represents the final velocity, \(u\) represents the initial velocity, \(a\) is acceleration, and \(d\) is displacement.
This equation is particularly useful when we know the initial and final velocities, as well as the distance traveled, and wish to find the acceleration.
For example, in the case of an electron accelerating in a TV tube, we are given:

• Initial speed, \(u = 0\) m/s.
• Final speed, \(v = 3.00 \times 10^6\) m/s.
• Distance, \(d = 0.018\) m.

By substituting these values, we determine the acceleration of the electron is approximately \(2.5 \times 10^{14}\) m/s².

This powerful equation allows us to understand how quickly an object speeds up over a certain distance.

Newton's second law

Newton's Second Law of Motion provides a fundamental principle for analyzing forces acting on an object. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:

• \[ F = ma \]

In this formula, \(F\) stands for force, \(m\) is mass, and \(a\) is acceleration.
This law helps us understand how different forces affect the motion of objects by linking force directly with mass and acceleration.

For the electron traveling through the TV tube, we use its mass, \(9.11 \times 10^{-31}\) kg, and calculated acceleration, \(2.5 \times 10^{14}\) m/s². Applying Newton's second law, we calculate the net force exerted on the electron:

• \[ F = 9.11 \times 10^{-31} \times 2.5 \times 10^{14} \approx 2.28 \times 10^{-16} \, \text{N} \]

This demonstrates how even particles with extremely small mass, like electrons, can experience significant forces when accelerated to high speeds.

electron dynamics

Electron dynamics covers the study of motion and forces acting on electrons. Electrons are fundamental particles with a small mass and negative charge. Their behavior is crucial in many technologies, like cathode ray tubes used in older televisions.
Understanding these dynamics involves looking at how electrons respond to forces, often in the context of fields or accelerators. For the electron in the TV tube, its dynamics are dictated by the need for rapid acceleration over a short distance to produce images on the screen.

In analyzing electron dynamics:

• We consider the initial conditions such as starting from rest.
• We calculate significant speed changes due to small forces over mini distances.
• We employ fundamental physics equations to find specific outcomes, like speed or force.

By understanding these principles, engineers design and improve electronic devices to use electron motion efficiently. This understanding allows electrons to be manipulated in necessary directions and speeds, optimally controlling their dynamic behavior in various applications.