Conceptual Questions (5 points each)

Questions 1 -- 3 refer to the following situation.

1. Which depiction below is a valid representation of the electric field lines of the two charges?

   

2. At which of the marked points can the electric field equal zero?

   B.

   The two contributions to the electric field, due to the -q and -4q charges, must cancel; that is, they must be equal in magnitude and opposite in direction. Both charges have an attractive field; thus to the left of the -q charge, both point to the right; to the right of the -4q charge, both point to the left. Only in between are the two electric field contributions in opposite directions, with the electric field due to the -q charge pointing left, and that due to the -4q charge pointing right. To set the magnitudes of the two contributions equal in this region, we note that the electric field due to Q is proportional to , where r is the distance from Q to where we are calculating the electric field. Thus to balance the electric field due to -q a distance d to the right of -q, we must be a distance 2d to the left of the -4q charge, to get the same magnitude . This means we must be closer to the weaker charge -q.

3. At point E, two uncharged conducting spheres, L and R, are suspended from insulating threads so that they touch each other, as shown. When the two spheres are separated, how will they be charged?

   

   

   While in contact, charge moves between the conducting spheres,with negative charges migrating to R due to the repulsive -4q charge, leaving an opposite positive charge on L. When L and R are separated, this charge separation persists, leaving L positive and R negative.

   Questions 4 -- 9 refer to the following situation:

   
   

   

4. What electric force is exerted on a -4 test charge placed just inside the top plate?

   (a) 0.6 N, upward

   , so that the electric force has magnitude . Because the test charge is negative, points opposite ; that is, upward.

5. What is the charge on the top plate of the capacitor?

   (c) C

   A capacitor has plates with charge +Q and -Q, where , which here equals C. To figure out the sign of charge on the top plate specifically, note that the electric field lines start on the top plate, meaning that it must have positive charge.

6. Which of the following is a valid plot of the equipotential curves between the capacitor plates?

   

7. How much work is required to move a -4 test charge horizontally from A to B, a distance of 2 mm?

   (e) zero

   Horizontal motion is along a single equipotential, perpendicular to the electric force; hence no work is done by the electric field. This is true (work done is zero) whenever the net change in potential of the motion is zero.

8. How much work is done by the electric field in moving a -4 test charge from the top to the bottom plate of the capacitor?

   (b) J

   Work done by the electric field is . Here we have a decrease in potential from top to bottom of 3000 V, giving J. Note that negative work makes sense, as the electron moves opposite the direction that it is pushed by the electric field.

9. Suppose that a dielectric sheet is inserted to completely fill the space between the plates, and the potential difference between the plates drops to 1000 V. What is the capacitance of the system after the dielectric is inserted?

   (e) F

   Inserting a dielectric increases the capacitance, with the new C given by . For fixed charge on the plates (since our capacitor is isolated), this reduces the voltage across the plates to , since always. Here is reduced by a factor of 3, so , and the new capacitance must be 3 times the old.

   Questions 10 -- 12 refer to the following circuit:

   
   

   

10. is measured to be 1.5 A. Which of the following gives the currents through the 10 and 20 resistors?

    (b)

    divides at the junction with the parallel 10 and 20 resistors, with its two branches, and , summing to the original current . Since the voltage drops across the parallel resistors are equal, each resistor draws a current which is inversely proportional to its resistance. Thus the 20 resistor, with twice the resistance, draws half as much current as the 10 resistor. (b) is the only choice satisfying both the correct proportionality and the correct sum.

11. Which of the following statements is correct, concerning the voltage drops , , and across the 10 , 20 and 30 resistors, respectively?

    (c) is greater than which equals .

    and are equal, because the 10 and 20 resistors are in parallel. Note that the current through the 30 resistor is greater than that of either parallel resistor (this because ; the 30 resistor sees the full current through both parallel resistors) Since it has the greatest current and the greatest resistance, the 30 resistor has the greates voltage drop .

12. What is the equivalent resistance of the 10 and 20 resistors?

    (c) 6.7

    Since they are in parallel,

    

Quantitative Problems (Point Value as Marked)

1. (24 points)
   Two charges of magnitude Q= 2.0 C are held 2.0 m apart as shown in the figure.

   

   1. What is the electric potential energy of the two charges; that is, what work is required to bring them from infinite separation to their current positions?
   2. What is the electric potential due to the charges at the point P shown?
   3. In what direction does the electric field point at P? SHOW the vector sum of electric fields due to both charges and state why they sum to this direction.

   
   
   
   
   
   
   
   
   

   1. We calculate the work required to assemble the charges. In bringing in the first charge , no work is done, because no electric field acts on the charge. In bringing in the second charge , however, we must do work, as we must overcome a repulsive electric force due to the first charge. That work is

      

      since the electric potential energy of is zero when infinitely far from . This work required is thus

      

      The electric potential energy of the charges, which is just the work required to assemble them, is thus

   2. The electric potential at P is just the sum of the electric potential at P due to each charge. Individually, this is just V= kq/r, where q is the charge and r is the distance from the charge to P. Thus

      

   3. Both charges exert a repulsive electric field, directed away from themselves as shown. Both electric fields have equal magnitude at P, and both make the same angle with the x-axis. Thus both have equal and opposite vertical components, and their vector sum points to the right.

      

2. (16 points)
   Consider the following circuit.

   
   
   
   
   
   
   
   
   
   
   
   
   

   Potential drops across each resistor are labeled; the currents shown flow from higher (+) to lower (-) potential. I have labeled my loops as shown. Choosing opposite directions for a loop will just multiply the entire loop equation by a sign. Kirchhoff's loop laws give, for each loop

   

   Two loop equations suffice to find and . Here I use loop 1 and loop 3, which yield

   

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