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In a television picture tube (CRT), electrons are accelerated by thousands of volts through a vacuum. If a television set is laid on its back, would electrons be able to move upward against the force of gravity? What potential difference, acting over a distance of \(3.5 \mathrm{~cm},\) would be needed to balance the downward force of gravity so that an electron would remain stationary? Assume that the electric field is uniform.

Short Answer

Expert verified
The potential difference required is approximately \(1.95 \times 10^{-12} \mathrm{~V}\).

Step by step solution

01

Define the Requirements

We need to determine the potential difference required for an electron to overcome gravitational force over a distance of 3.5 cm. This involves equating the electric force to the gravitational force acting on the electron.
02

Identify Forces Acting on the Electron

The gravitational force acting on an electron is given by the equation \( F_g = m imes g \), where \( m = 9.11 \times 10^{-31} \) kg is the electron's mass and \( g = 9.8 \) m/s² is the acceleration due to gravity. The electric force can be expressed as \( F_e = e \times E \), where \( e = 1.6 \times 10^{-19} \) C is the charge of an electron, and \( E \) is the electric field strength.
03

Equate Electric and Gravitational Forces

To balance the electron, set \( F_e = F_g \). This results in the equation: \( e \times E = m \times g \).Solving for \( E \) (electric field strength), we find:\( E = \frac{m \times g}{e} \).
04

Calculate the Electric Field Strength

Substitute the known values into the formula:\( E = \frac{9.11 \times 10^{-31} \mathrm{~kg} \times 9.8 \mathrm{~m/s^2}}{1.6 \times 10^{-19} \mathrm{~C}} \).Evaluate this to find \( E \approx 5.58 \times 10^{-11} \mathrm{~N/C} \).
05

Find the Required Potential Difference

To find the potential difference (\( V \)), use the relation between electric field and potential difference: \( V = E \times d \), where \( d = 3.5 \) cm (or 0.035 m). Plug in the values to calculate:\( V = 5.58 \times 10^{-11} \mathrm{~N/C} \times 0.035 \mathrm{~m} \).
06

Evaluate the Potential Difference

Evaluate the expression to find the potential difference:\( V \approx 1.95 \times 10^{-12} \mathrm{~V} \).This is the potential difference needed for the electron to remain stationary against gravity.

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Key Concepts

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

Potential Difference
The potential difference, also known as voltage, is a measure of the work needed to move a charge from one point to another in an electric field. It is essentially the energy required per unit charge to move across a certain distance.
In practical terms, think of potential difference as the electric 'pressure' that pushes charges through an electric field.
  • It's measured in volts (V).
  • One volt is equivalent to one joule per coulomb.
The potential difference is crucial in many devices, like cathode ray tubes (CRTs), where it accelerates electrons across the tube.

In our exercise, we use potential difference to generate an electric force strong enough to counteract the gravitational force on an electron. This balance allows it to remain stationary instead of succumbing to gravity.
Gravitational Force
Gravitational force is a fundamental force that attracts any objects with mass towards each other. It's most noticeable when we consider the weight of objects on Earth's surface.
  • This force can be calculated using the formula: \( F_g = m \times g \).
  • Here, \( m \) is the mass of an object and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.
For an electron, although it is incredibly small with a mass of \( 9.11 \times 10^{-31} \text{ kg} \), it still experiences gravitational force.

When a force, like gravity, acts on an electron, it naturally accelerates in the direction of the force unless counteracted by another force, like an electric field.
Electron Acceleration
In a CRT, electrons are accelerated using electric potential difference. Acceleration is the rate of change of velocity of an object and is influenced by forces acting on that object.
For electrons, the acceleration involves both the gravitational force and the electric force created by the potential difference.
  • Electrons accelerate faster towards positively charged areas because they are negatively charged themselves.
  • Their acceleration can be quantified using \( F = m \times a \), where \( F \) is the net force acting on the electron.
This equation is crucial for understanding how CRTs work, as it helps us calculate how much voltage is needed to fire electrons towards the screen, forming images.

In our problem, the potential difference creates an electric field strong enough to make the electron's net acceleration zero, keeping it stationary.
Coulomb's Law
Coulomb's Law is a principle describing the force between two charges.
It states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
  • This law mathematically is: \( F = k \frac{|q_1 \times q_2|}{r^2} \).
  • Where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are charges, and \( r \) is the distance between the charges.
Understanding this law is fundamental in physics because it helps explain how electric forces are calculated, which is essential in dealing with electrons in CRTs.

In our context, though Coulomb's Law isn't directly used to calculate the potential difference, its principle underlies the interactions within the electric field, affecting how strongly the field can push the electrons against gravity.

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Most popular questions from this chapter

(III) You are trying to determine an unknown amount of charge using only a voltmeter and a ruler, knowing that it is either a single sheet of charge or a point charge that is creating it. You determine the direction of greatest change of potential, and then measure potentials along a line in that direction. The potential versus position (note that the zero of position is arbitrary, and the potential is measured relative to ground) is measured as follows: $$ \begin{array}{lllllllllll} \hline x(\mathrm{~cm}) & 0.0 & 1.0 & 2.0 & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 & 8.0 & 9.0 \\ \mathrm{~V} \text { (volts) } & 3.9 & 3.0 & 2.5 & 2.0 & 1.7 & 1.5 & 1.4 & 1.4 & 1.2 & 1.1 \\ \hline \end{array} $$ (a) Graph V versus position. Do you think the field is caused by a sheet or a point charge? \((b)\) Graph the data in such a way that you can determine the magnitude of the charge and determine that value. \((c)\) Is it possible to determine where the charge is from this data? If so, give the position of the charge.

(II) A manufacturer claims that a carpet will not generate more than \(5.0 \mathrm{kV}\) of static electricity. What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a \(5.0-\mathrm{kV}\) potential difference between the shoe and the carpet, approximating the shoe and the carpet as large sheets of charge separated by a distance \(d=1.0 \mathrm{~mm} ?\)

(II) A total charge \(Q\) is uniformly distributed on a thread of length \(\ell\). The thread forms a semicircle. What is the potential at the center? (Assume \(V=0\) at large distances.)

(I) How much work does the electric field do in moving a proton from a point with a potential of \(+185 \mathrm{~V}\) to a point where it is \(-55 \mathrm{~V} ?\)

(III) Electrons are accelerated by \(6.0 \mathrm{kV}\) in a CRT. The screen is \(28 \mathrm{~cm}\) wide and is \(34 \mathrm{~cm}\) from the \(2.6-\mathrm{cm}\) -long deflection plates. Over what range must the horizontally deflecting electric field vary to sweep the beam fully across the screen?

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