7 because the plates are pulled apart after the

This preview shows 1 out of 3 pages.

7. Because the plates are pulled apart after the capacitor was isolated, the charge of 3.0 C will remain constant. We can use that and the new capacitance to find the change in energy stored (which will be the amount of work done on the capacitor) by using the U = Q 2 / (2 C ) version of energy stored by a capacitor. We’ll get D. 8. Combine the three resistors on the left (1 , 1 and 2 ) in series to get a 4 resistor. This is in parallel with the original 4 resistor so they combine to make a 2 resistor. This new 2 resistor is in series with the remaining original 1 and 2 resistors so they combine to make a 5 resistor. This is in parallel with the last remaining original resistor (5 ) so the equivalent resistance is 2.5 (answer D).
Image of page 1

Subscribe to view the full document.

9. From Ohm’s Law ( R V I ), the wire with twice the current must have half the resist ance. That makes wire 1’s resistance R/2 (answer B). 10. From d V E , the electric field is strongest where the equipotential lines are closest together so E 2 is biggest (eliminates all but C and E) and E 3 is smallest (making C correct). 11. The potential energy will decrease as the positive charge gets closer to the negative side (which is the same as saying the potential energy gets smaller as the positive charge gets farther from the positive side). This is to the left (B). 12. Since the object is suspended, the net force acting on it is zero. Since its weight is 49 N, and the tension is only 15 N, the electric force must point up. The electric field points up so the electric force is in the same direction as the field and the charge must be positive (eliminating A C). The magnitude of the force must be the difference between the magnitudes of the weight and tension (34 N). From the definition of electric field, ( q F E ), the charge is 52 mC (answer D).
Image of page 2
Image of page 3
You've reached the end of this preview.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern