1. + An electron and a proton are set free, near each other, deep in outer space. The electron moves towards the proton with 1) constant velocity ( v  constant ) 2) increasing velocity but constant acceleration ( a  constant ) 3) increasing velocity and increasing acceleration

2. + An electron and a proton are set free, near each other, deep in outer space. The electron moves towards the proton with? There is a net force on the electron due to the proton’s charge, so the electron accelerates (1 is out). As the electron moves closer to the proton, the force it experiences grows stronger (Coulomb’s Law holds F  1/r 2 ). If the force becomes stronger, and the mass does not change, then Newton’s Second Law (F = ma or a = F/m) says that the acceleration increases. The answer must be (3).

3. + An electron and a proton are set free, near each other, deep in outer space. What about the proton?

4. m proton = 0.000 000 000 000 000 000 000 000 001 6748 kg m electron =0.000 000 000 000 000 000 000 000 000 0009 kg m proton = 1 e  q e =1.60210  10  19 Coulomb 1836  m electron

5. How many electrons are there in 1 Coulomb of negative charge? How many protons are there in 1 Coulomb of positive charge?

6. Two uniformly charged spheres are firmly fastened by insulated stands to frictionless pucks which are set on an air table. The charge on sphere 2 is 3  the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces? + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 1 4 2 3

7. What force exists between two concentrations of charge, each 1 coulomb separated center-to-center by 1 meter? ( 1 C )( 1 C ) (1 m 2 ) = 2,020,390,000 lbs = 1,010,195 tons  0.2248 lb/N = 8.9875  10 9 N

8.  = 15 o assume 2  equally charged balloons (with uniformly distributed charge)  0.8 grams each W  0.008 N 0.75 meters

9.  0.75 m x 1) 0.75sin  2) 0.75cos  3) 0.75/sin  4) 0.75/cos  5) sin  /0.75 6) sin  /0.75 We can calculate the distance, x, by

10.  0.75 m x

11. BALLOON’S WEIGHT STRING’S TENSION ELECTROSTATIC REPULSION 

12. W T F  T F W

13. W T F electric  We can calculate the tan  by 1) T/F 2) T/W 3) F/T 4) F/W 5) W/F 6) W/T

14. W T F electric  With tan 15 o = F electric W F electric =(0.008 N)  tan15 o = 2.144  10 -3 N

15. W  0.008 N R = 0.388 meters 0.002144 N = (8.9875  10 9 Nm 2 /C 2 ) q 1 q 2 (0.388 m) 2 q 1 q 2 = (0.388 m) 2  0.002144 N / (8.9875  10 9 Nm 2 /C) = 3.59  10 -14 C 2 Assuming q 1  q 2, that means each q  1.89  10 -7 Coulomb How many electrons would that be?