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.
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
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This book is written by our very own Professor Fullerton and features more than 600 worked-out problems with full solutions and deeper understanding questions. AP Physics 1 Essentials covers all major topics included in the AP Physics 1 course, including: kinematics, dynamics, momentum, impulse, gravity, uniform circular motion, rotation, work, energy, power, mechanical waves, sound, electrostatics, and circuits.