MT2 S2 Explanations

# 7 because the plates are pulled apart after the

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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).

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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).
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