Physics 9 Friday, February 21, 2014

Transcription

1 Physics 9 Friday, February 21, 2014 Turn in HW5 in class today. Next week: just one chapter (Ch. 24: Gauss s law). Read first half (concepts) for Monday & answer online Q s. Today: some demos and a worksheet (!) a propos electric forces and the electric field.

2 Electrostatic forces i.e. forces due to electric charges that are not in motion The electrostatic force due to b acting ON a is F electric b ON a = k q a q b r 2 ab ˆr b a (or in the book s more abbreviated notation) F E ba = k q a q b r 2 ab ˆr ba where k = N m 2 /C 2. The force ON a points away from b for like charges and points toward b for opposite charges. (Opposites attract, likes repel.) (Ask yourself how you would rewrite this equation to get the force due to a acting on b.)

3 The net force ON a due to a set of N charged objects is F E a = k N i=1 q i q a r 2 ia where r ia is the distance from object i to object a and ˆr ia is the unit vector pointing from object i toward object a. ˆr ia Math reminder: the unit vector pointing in direction r = (x, y, z) is ˆr = r r = (x, y, z) x 2 + y 2 + z 2 So ˆr points in the same direction as r, but ˆr has a length of 1.

4 That brings us to worksheet Q#1. (Work on it, then we ll vote.)

5 The magnitude of the electric force FAB E exerted by A on B is (A) N (B) N (C) 0.45 N (D) 0.90 N (E) 4.5 N

6 The magnitude of the electric force FBA E exerted by B on A is (A) N (B) N (C) 0.45 N (D) 0.90 N (E) 4.5 N

7 The magnitude of the electric force FAC E exerted by A on C is (A) N (B) N (C) 0.45 N (D) 0.90 N (E) 4.5 N

8 The magnitude of the electric force FBC E exerted by B on C is (A) N (B) N (C) 0.45 N (D) 0.90 N (E) 4.5 N

9 My answers (A) did (B) did not look something like this.

10 (A) Question (1) is too easy. Not a good use of classroom time. (B) I understand question (1) now. This was helpful. (C) I still don t understand how to do question 1. (D) None of the above.

11 (2) Draw arrows for the two electric forces that are acting ON particle C. (The electric force exerted by A on C is written F AC E. The electric force exerted by B on C is written F BC E.) Then draw an arrow for the vector sum of forces (a.k.a. the net force ) acting on particle C, which is written F E C. To make it easier to compare results, choose the length of your arrows so that the grid size on your force diagram is 0.1 N. (Use the left grid below.)

12 (2) Draw arrows for the two electric forces that are acting ON particle C. (The electric force exerted by A on C is written F AC E. The electric force exerted by B on C is written F BC E.) Then draw an arrow for the vector sum of forces (a.k.a. the net force ) acting on particle C, which is written F E C. To make it easier to compare results, choose the length of your arrows so that the grid size on your force diagram is 0.1 N. (Use the left grid below.) My answer (A) did (B) did not look something like this.

13 (3) Now draw (above right) arrows for the forces acting on particle B and their vector sum. Again use a grid size of 0.1 N.

14 My answer (A) did (B) did not look something like this. (3) Now draw (above right) arrows for the forces acting on particle B and their vector sum. Again use a grid size of 0.1 N.

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17 Electric field (E) E(x, y, z) is force-per-unit-charge that a test charge q, if placed at position r = (x, y, z), would feel as a result of the other charges. If we put an object of charge Q at the origin, the force on q is F q = k Qq r 2 Qq ˆr Qq = +k qq r 2 ˆr So the electric field E( r) is E(x, y, z) = F on q q = k Q r 2 ˆr The magnitude of E falls off like 1/r 2, and (for positive Q) E points away from Q. E points away from positive source charges and points toward negative source charges.

18 Electric field due to N charged objects Q 1, Q 2,..., Q N Just as the force experienced by a test charge q (positioned at point P) is the vector sum of the forces due to the other charges, the electric field E (evaluated at point P) due to N charged objects is the vector sum of the contributions from each charge: E(P) = F on q q = k N i=1 Q i r 2 ip That implies that the electric field due to N different source objects is just the superposition (i.e. the vector sum) of the electric fields due to the individual sources. ˆr ip

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23 Triboelectric series (who knew?!) most positively charged air human skin (dry) leather rabbit fur glass quartz mica human hair nylon wool lead cat fur silk aluminum paper (small positive charge) (small negative charge) wood rubber balloon resins hard rubber nickel, copper brass, silver gold, platinum synthetic rubber polyester styrofoam plastic wrap scotch tape vinyl teflon silicone rubber most negatively charged

24 Physics 9 Friday, February 21, 2014 Turn in HW5 in class today. Next week: just one chapter (Ch. 24: Gauss s law). Read first half (concepts) for Monday & answer online Q s. Today: some demos and a worksheet (!) a propos electric forces and the electric field.