Electric Fields Practice Problems

Electric Fields Practice Problems
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  • 0:03 Electric Field
  • 1:10 Single Electric Field
  • 2:25 Superposition of…
  • 3:57 Electric Field in…
  • 5:34 Faraday Cages
  • 6:05 Lesson Summary
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Lesson Transcript
Instructor: Michael Blosser

Michael has a Masters in Physics and a Masters in International Development. He has over 5 years of teaching experience, teaching Physics, Math, and English classes.

In this lesson, you will be introduced to electric fields so that you can practice calculating single fields, multiple fields, and electric fields in charged spheres. You'll also learn how Faraday cages work.

Electric Field

An electric field is created by any charged object and is defined by the electric force divided by the unit charge. We can also define an electric field with this equation:

E = F/q

Where:

  • E is the electric field in Newtons/Coulombs
  • F is the electric force in Newtons
  • q is the charge in Coulombs

Or if we are doing with point charges, we can use Coulomb's law to get this equation for an electric field of point charges:

Electric field equation

where:

  • q is the charge in Coulombs
  • r is the distance away from the point charge
  • k is the electric force constant of 9 * 109 N * m2 / C-2

Electric fields act in a similar way to gravitation fields, in that the electric forces created by an electric field can act at a distance. There are four standard electric field problems or examples that will help us review the properties of electric fields and how to calculate them. Let's look at them now.

Single Electric Field

Our first example involves a single electric field. Consider the following questions:

  • What is the electric field 10 meters away from a point charge that has a charge of 10 Coulombs?
  • What would be the electric field strength at a distance of 1 nanometer?

Answer

First, we need to determine what equation to use to calculate the electric field strength. We've been given the equation for the electric field of point charges. Since we're dealing with point charges, that's the equation we will use. Therefore, we can solve for the electric field strength at the two distances given by doing the following:

From 10 meters away:

Electric field equation

Pt charge equation 1

This gives us an answer of an electric field strength of 9 * 108 N/C for a distance of 10 meters away.

Now, from 1 nanometer away:

Electric field equation

pt charge 2

This gives us an answer of an electric field strength of 9 * 1028 N/C for a distance of 1 nanometer.

Superposition of Electric Fields

Now, let's look at an example involving superposition of electric fields:

A point charge q1 of 10 microcoulombs is a horizontal distance d of 1 cm to the left of a point charge q2 of 1 microcoulombs. What would be the electric field strength of a distance x that is 2 cm to the right of point charge q2?

Answer

In order to solve this equation, we need to use the principle of superposition. This principle tells us that the electric field from any number of point charges is the vector sum of all the point charges. This can be represented with the following equation:

Etotal = E1 + E2 + E3 and so on…

Therefore, we must add up the electric field of each charge at each individual charge's distance away from point x. This gives us the equation of:

superposition equation

superpostion equation with variables

Plugging in our variables gives us the electric field strength at point x of 1.225 * 108 N/C.

Electric Field in Charged Sphere

Next, let's look at an example involving a charged sphere:

  • What would be the electric field strength a distance of 1 cm from a charged sphere of 10 microcoulombs?
  • What would be the electric field strength inside the sphere?

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