Quiz 1 Ch. 21 ~ MP1, 82, 8, 10, 22, 76, 36, 34, 33, 35, MP2, 47, 54, 53, 55, MP3, MP4, 60, 59  
EVEN ANSWERS
HYPERPHYSICS
THIS SIMULATION  will illustrate principles of Electric forces and fields; it's related to #33 and 35 below.  

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.
Some problem are titled"MP, numbered make up problems with hints,  ie. MP1, MP2, MP3, MP4 below.  While the MP's are not to be turned in , they could be included on the exam as required material which we will review for tests. 

(CHECK YOURSELF WITH  MADE UP PROBLEM1 (MP1):) (a) Assuming gravity is the only force acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a free falling object at the Earth's surface.
(b) Suppose the earth were made only of protons but had the same size and mass as it presently has? What would be the acceleration of an electron released at the surface? Is it necessary to consider the gravitational attraction as as well as the electric force ? Why or why not? 
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(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.

82. YOU CANNOT IGNORE MUTUAL ATTRACTION OF SPHERES, WHICH INCREASES THE ELECTRIC  FIELD MAGNITUDE E BETWEEN PLATES. 
 (a) Clearly the ball on the right is negative. EXPLAIN.
(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 25  + k*q2/(4L2sin225)
y-DIRECTION: m*g = T*cos 25.

SOLVING THESE EQUATIONS SIMULTANEOUSLY for E gives E = 3660 N/C.

If you do not take into account the mutual attraction your answer is logically smaller E = 432 N/C. 

 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)}
10. See class notes 1-22 -14 about the bricks(blobs)

22. 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 smaller than the leftmost charge.  See example 3 for guidance though the problem is different.  

76. The two unknown charges have equal magnitude but opposite signs; q1 is negative.  
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 8. 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 21.7 . 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.

 MP2. Example 7.  
(a) For the electron in example 7, compare the weight of the electron with the magnitude of the electric force on the electron. Give a ratio.  It is appropriate to ignore the gravitational force in these examples? EXPLAIN.  
(b) A particle with charge +e is placed a rest between the charged plates in figure 21.20. What must of the mass of this object be if it is to remain at rest? Give your answer in kg and in multiples of the electron mass. (c) Does the answer to (b) depend on where between the plates the object is placed? Why or why not?  
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(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 8. 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 10.
(b) Use formula 21.8

53.  (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.  

55.  
(a) Formula 21.11 
(b) Formula 21.8
(c) Would behave like a point charge.
(d) Think !
MP3. 
Two Horizontal, Infinite plane sheets of charge are separated by a distance d. The lower sheet has negative charge with uniform surface charge density  -rho <0. The upper sheet has positive charge with uniform surface charge density  rho > 0. What is the electric field magnitude and direction 

(a) above the sheet 
(b) below the sheet 
(c) between the sheets? 
Use the symbol rho as needed. 
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Example 12. 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 

MP4. Sketch the electric field lines for a disk of radius R with a uniform surface charge density rho.  Use what you know about the electric field very close to the disc and very far from the disk to make your sketch.   
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 See example 11 in the limit of very large R, example 12, and also Figure 21.28 . 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.

 60. See example 14. Evaluate on x-axis now.
Hint: The direction  of the electric field is vertically down.  

 59. (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. 21.28 (b) and compare with the direction of the external field.