Quiz 1 Ch. 21 ~ 10, 84, 8, 20, 78, 36, 34, 33, 35, 37, 47, 54, 55, 56, 57, 60, 65, 69, 84 

DISCUSSIONS; This quiz is example driven . READ ALL EXAMPLES !  In the following discussions,  I will also  often reference  corresponding example  giving  you a solid hint for get shift that car out of park and on the road.

10. See example 1.  Below, m is the electron mass and e is the electron charge magnitude.
(a) mg = k*e2/D2. Solve for D. (b)  Get the number of protons in the Earth's mass: ME/mp = n and find m*a = k*n*e2/RE2.  Find acceleration a.

84.  (a) Clearly the ball on the right is negative.
(b)  Take the ball on the  left which is positive. Break the forces into  x and y components and use equilibrium---
x-DIRECTION: q*E = T*sin theta/2
y-DIRECTION: m*g = T*cos theta/2.

SOLVE THESE EQUATIONS SIMULTANEOUSLY  for theta. 

8. (a) number of electrons = (atomic number)* (number of moles)*(avogadro's number), where number of moles = mass/atomic mass.
(b) force = 10,000 N = k*q2/D2.  Solve for q then divide by electron charge magnitude e.
(c) {compute answer to (b)}/ {answer to (a)}

20. See example 3. The 5.00 nC charge is 0.500 m to the left of the 4.00 nC charge.  The middle   negative charge is 0.300 m  to the right of the leftmost charge.  Net force is points  RIGHT from Coulomb's Law since the middle charge is closer to the rightmost charge even though the magnitude of that right charge is the smallest of the three. See example 3 for guidance though the problem is different.  

78. The two unknown charges have equal magnitude but opposite signs; q1 is negative.   See ICQ1 and variations. 
mass*acceleration = Net Force = 2*F*cos theta , upward , where cos theta = 2.25/3.00 and F = k*|Q|*q/(0.003)2   .  Find q= magnitude of the unknown two charges.

 36.  (a) See example 6. Use the Pythagorean Theorem and find the related angle by computing the ratio of the absolute values of the y and x components of the Field. (b) (i) The force will be opposite the electric field at point P example 6. (ii) The force will be  opposite the force in part (i)

 34. See examples 6 and 9. A   3-4-5 triangle is formed by point P and the two charges. The hypotenuse is between point P and q2. The short side is between point P and q1 at origin  (a) The electric field due to charge 1 points vertically down with magnitude k* |q1|/(0.04 m)2.  The electric field due to charge two makes an angle 53 degrees with  the negative x axis and has magnitude k* |q2|/(0.05 m)2  .

 33. (a) See example 21.8 . It is reasonable to ignore gravity.  See also the SIMULATION.

change in y = (1/2)*(e*E/m)*t2 and L = Vx*t, where the x-component of velocity, L and change in y are given; find E= field magnitude. 
(b) In the case of a proton , you are given the proton mass. Plug this into the two above equations and solve for change in y assuming the same L and time t.

(c ) and (d) use info from the above parts.

 35. speed2 = Vx2  + Vy2. , where the x-component of velocity is known. And the y-component of velocity is  (eE/m)*t, where t is discussed in part (a)  of problem 33.

 37. Example 8.

(a) Compare m*g and e*E.
(b) Set m*g = e*E
(c) No, as long as you are well away from the ends.

 47. Example 9. The net electric field points left and has magnitude E ‘ + 2*E*cos 53, where E’ =
k*(2 uC)/(0.06)2  and

E = k*(5 uC)/(0.10) 2.  and uC means micro-Coulomb.

 54. (a) Use formula (equation) 21.9 and 21.10. Note as a becomes much larger than x, the formula becomes the equation for an infinitely long line—See example 11.
(b) Use formula 21.8

55.  (a) See formula 21.8.  Field points right.
(b) The force on the ring would be attractive and points right and it  can be found using Newton ’s 3Rd Law of action and reaction.

56.  (a) Formula 21.11 (b) Formula 21.8
(c) Would behave like a point charge.
(d) Think !


 57. Example 13. See my lecture notes when I discussed the field between the plates of a parallel-plate capacitor. You should be able to prove that above  top sheet and below the bottom sheet the net field due to both sheets is ZERO.  Between the top and bottom sheet, the field is twice the  magnitude of one sheet and points vertically downward.

60. See example 12 in the limit of very large R, example 13, and also Figure
21. 29 . Near the disk surface and away from edges , the system is like that of  an infinite plane. Far away from the disk, it looks like  a point charge.

 65.  Manipulate carefully; will review in class. 

 69. (a) the torque is zero when the angle  = 0 and the angle  = 180 degrees.
(b) The dipole tends to rotate so it points in the same direction as the external field, so clearly the zero angle is the position of stability.
(c) Draw the field lines of the dipole as shown in fig. 21. 29(b) and compare with the direction of the external field.