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 like charges repel

Electric Charge and The Electric Field. Static Electricity (Demo). There are two kinds of charge. -. +. Properties of charges.  like charges repel.  un like charges attract.  charges can move but charge is conserved.

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 like charges repel

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  1. Electric Charge and The Electric Field Static Electricity (Demo) There are two kinds of charge. - + Properties of charges  like charges repel  unlike charges attract  charges can move but charge is conserved Law of conservation of charge: the net amount of electric charge produced in any process is zero.

  2. Although there are two kinds of charge in an atom, electrons are the charges that usually move around. A proton is roughly 2000 times more massive than an electron. The charge on an electron is 1.6x10-19 coulombs. Conductors and insulators There is no such thing as a perfect conductor. A superconductor has no resistance to the flow of current, but is not the same as a perfect conductor. There is also no such thing as a perfect insulator.

  3. Induced Charge: The Electroscope Demonstrate (if electroscope available). Coulomb’s Law We’ve seen attractive and repulsive electrical forces at work. Coulomb’s law quantifies the magnitude of the electrostatic force. Coulomb’s law gives the force between charges Q1 and Q2 (in coulombs), where r is the distance in meters between the charges, and k=9x109 N•m2/C2. attractive for unlike OSE:

  4. Force is a vector quantity, so the equation by itself gives the magnitude of the force. Note how the direction is specified. If the charges are opposite in sign, the force is attractive; if the charges are the same in sign, the force is repulsive. Also, the constant k is equal to 1/40, where 0=8.85x10-12 C2/N• m2. To make it into a “really good” OSE I should specify “repulsive for like,” but that makes it too wordy. You’ll just have to remember that! attractive for unlike OSE: I have to do it this way for my Physics 35 class, the algebra-based second semester of College Physics. Later I’ll show a better way to write it.

  5. The equation as written is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed, r is the distance between the centers of the spheres. If more than one charge is involved, the net force is the vector sum of all forces (let’s not make that a required OSE). For objects with complex shapes, you must add up all the forces acting on each separate charge (calculus!). Note that we could have agreed that in this formula, the symbols Q1 and Q2 stand for the magnitudes of the charges. In that case, the absolute value signs would be unnecessary. We’ll do that in later OSE’s, but for now I want to emphasize the magnitude part. On your homework diagrams, show both the magnitudes and signs of Q1 and Q2.

  6. y Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm Solving Problems Involving Coulomb’s Law and Vectors You may wish to review vectors. Example: Calculate the net electrostatic force on charge Q3 due to the charges Q1 and Q2.

  7. Step 0: Think! This is a Coulomb’s Law problem (all we have to work with, so far). We only want the forces on Q3. Don’t worry about other forces. Forces are additive, so we can calculate F32 and F31 and add the two. If we do our vector addition using components, we must resolve our forces into their x- and y-components.

  8. F32 F31 y Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm Step 1: Diagram Draw a representative sketch—done. Draw and label relevant quantities—done. Draw axes, showing origin and directions—done. Draw and label forces (only those on Q3). Draw components of forces which are not along axes.

  9. F32 F31 y Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm Step 2: OSE <complaining> ”Do I have to put in the absolute value signs?” Yes, for now.

  10. F32 F31 y Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm Step 3: Replace Generic Quantities by Specifics (from diagram) Can you put numbers in at this point? OK for this problem. You would get F32,y = 330 N and F32,x = 0 N.

  11. F32 F31 y Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm Step 3 (continued) (+ sign comes from diagram) (- sign comes from diagram) Can you put numbers in at this point? OK for this problem. You would get F31,x = +120 N and F31,y = -70 N.

  12. F32 F3 F31 y Q3=+65C 30 cm 60 cm =30º x You know how to calculate the magnitude F3 and the angle between F3 and the x-axis. (If not, holler!) Q1=-86C Q2=+50C 52 cm Step 4: Complete the Math The net force is the vector sum of all the forces on Q3. F3x = F31,x + F32,x = 120 N + 0 N = 120 N F3y = F31,y + F32,y = -70 N + 330 N = 260 N

  13. If you define as the unit vector (a vector of unit length) pointing from Q1 to Q2, and F12 is the force on Q1 due to Q2, and you include the correct signs in Q1 and Q2, then - + Remember I said I’d show you a better way to write Coulomb’s law? I have to write like this because my College Physics students are not supposed to have to use unit vectors: attractive for unlike OSE: Q2 r12 Much more satisfying! Q1

  14. I did a sample Coulomb’s law calculation using three point charges. How do you apply Coulomb’s law to objects that contain distributions of charges? We need another tool to do that.

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