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Electronic technicians complete a number of tasks when assembling circuit boards.  A voltmeter is frequently used when testing circuits.  The word  voltage  stems from the fact that potential difference is measured in volts with a voltmeter.

 

In such instances, the voltage is describing the amount of work required per unit of charge to move a test charge from one point to another within a circuit.  Because the calculation involves subtracting one potential value from another, this quantity is called  electric potential difference  (or just potential difference for short).



electric potential difference:  the change in electric potential experienced by a charge moving from one point to another in an electric field.

 

Electric potential difference is sometimes called potential difference.


In terms of an equation, electric potential difference is described as

 

 

The following calculation shows how to calculate the electric potential difference if the test charge in the previous lab activity was moved from an initial location 300 m from the source to a final location 100 m from the source.

 

 

 

 


Read Unit Introduction
Read the sections called "Electric Potential" and "Electric Potential Difference" on pages 563 to 566 of the text.  Pay close attention to the explanation of the electron volt and to "Example 11.8." 


Self-Check

Answer the following self-check (SC) questions then click the "Check your work" bar to assess your responses.

 

SC 10.

An electron is accelerated to high velocities within the picture tube of an old-style CRT (cathode ray tube) computer monitor. The potential difference used in the picture tube is 375 V.

  1. The charge of an electron is 1.60 × 10 -19 C. Calculate the energy gained by the electron in joules.
  2. Calculate the energy gained by the electron in electron volts.
  3. Consider your answers to SC 10.a. and SC 10.b. Suggest a reason why the electron volt is considered to be a convenient unit for researchers studying the behaviour of electrons and other particles in electric fields.

SC 11.  

  1. Complete "Practice Problem" 2 on page 566 of your textbook.
  2. Use your answer from SC 11.a. to determine the final velocity of the electron assuming that it started at 8.00 × 10 7 m/s. Remember that you can find the mass of the electron on your data sheet. NOTE: The text asked you to determine the electric potential energy gained. There is no problem with this, however the answer to b assumes it was kinetic energy that was gained. So do the calculation on that basis and find the larger speed.
 

SC 10.

Given

 

Required

The gain in energy of the electrons.

 

Analysis and Solution

 

Paraphrase

The electrons gain 6.00 × 10 −17  J.


Method 1

 

Given

The energy was calculated as Joules in SC 10.a.

 

Δ E  = 6.00 × 10 −17  J

1eV = 1.60 × 10 −19  J

 

The conversion fraction is on the physics data sheet.

 

Required

The energy in electron volts.

 

Analysis and Solution

 

Paraphrase

The electron gains 375 eV.

 

Method 2

 

Given

Δ V  = 375 V

q  = 1 e

 

Required

The energy in electron volts.

 

Analysis and Solution

 

Note 1  e  (1 elementary charge) is an exact value, therefore does not count for significant digits.

 

Paraphrase

The electron gains 375 eV.

 

The answer to the previous question labelled "Method 2" simply requires multiplying the value of the potential difference in volts by the charge of 1 electron. In this case, if an electron is accelerated through 375 V the energy gained is 375 eV. What could be easier?


SC 11.

Given

Δ V  = 4.00×10 4  V

q  = 1.60×10 −19  C

 

Required

The energy of the electron in Joules and electron volts

 

Analysis and Solution

Since the electron has been accelerated by 4.00 × 10 4  V, the energy it has gained is 4.00×10 4  eV.

 

An energy value in Joules can be determined using the equation or by unit conversion.

 

Equation

 

Unit Conversion

 


 

Paraphrase

The electron gains 4.00 × 10 4  eV or 6.40 × 10 −15  J of energy.


Given

 

Required

The final velocity of the electron.

 

Analysis and Solution

Using the conservation of energy, the energy gained by the electron will be converted into kinetic energy.

 

 

Paraphrase

The final velocity of the electron is 1.43 × 10 8  m/s.