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### Electric Fields & Forces

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Question 1 0:19
• Question 2 0:51
• Question 3 1:30
• Question 4 2:19
• Question 5 3:12

### Transcription: Electric Fields & Forces

Hello and welcome back to Educator.com.0000

I am Dan Fullerton and in this lesson, we are going to go over page 1 of the APlusPhysics worksheet on electric field and you can find a link to that worksheet down below the video.0003

Take a minute, download it, print it out, see how you do and then we will use this video to check answers.0012

Number 1 -- Which diagram represents the electric field lines between two small electrically charged spheres?0019

Remember field lines run away from positives and into negatives.0025

Well, Number 1 cannot be in it, because we are going into a positive.0029

Number 2 -- we are going away from positives -- that looks good.0033

Number 3 we are going away from negatives and into positives -- that is not it and then number 4, we are also having the field lines run away from a positive, so that cannot be it.0037

Our best answer there, must be Number 2.0046

Number 2 -- Which graph best represents the relationship between the magnitude of the electric field strength (E) and a point charged in the distance (r) from the point charged?0051

Remember (E) electric field, is the electric force divided by charge, which is (k)(q1)(q2)/r2.0061

There is the force and we have to divided by that charge, so it is going to be (k)(q)/r2 -- still an inverse-square law relationship, so as distance gets bigger, we are going to go down by the square of the distance, so our correct answer here must be Number 4.0070

Number 3 -- The diagram below represents an electron within an electric field between two parallel plates that are charged with a potential difference of 40 volts.0090

If the magnitude of the electric force on the electron is 2 × 10-15 N, the magnitude of the electric field strength between the charged plates is...?0100

Well, to get the electric field strength, that is the electric force divided by the charge, which will be 2 × 10-15 N/1.6 × 10-19 coulombs (C) (charge on an electron)...0109

...and when I do that I come up with an answer around 1.25 × 104 N/C, so the correct answer there is Number 3.0125

Number 4 -- We have two oppositely charged parallel metal plates 1 cm apart and they exert a force with a magnitude of 3.6 × 10-15 N on an electron placed between the plates.0139

Calculate the magnitude of the electric field strength between the plates.0151

Well, the electric field strength is going to be the electric force divided by the charge, which is 3.6 × 10-15N/1.6 × 10-19 C (charge on an electron).0156

When I divide that using my calculator, I come up with something right around 22,500 N/C or 2.25 × 104 N/C.0175

One more here -- An electron is located in the electric field between two parallel metal plates as shown in this diagram. 0199 If the electron is attracted to Plate (A), then Plate (A) is charged.0190

Well, if the electron is attracted to Plate (A), that must be the positive potential and (B) must be the negative side.0203

We also know the electric field runs from positive to negative.0211

The electron fields are forced in the opposite direction of the electric field, so it must be going up.0217

Having figured that out, let us pick our best answer here.0222

Plate (A) is charged -- well positively -- so we can get rid of answers 3 and 4 and the electric field is directed from (A) to (B).0225

Number 1 there must be our best answer.0234

All right. If these went pretty well -- Great -- You are probably ready to try the AP problems.0238

If they did not go so well, now is a great time to go back and look over our full lesson on electric fields.0242

Thanks so much for your time everyone and make it a great day!0249