This exam consists of 31 questions; true-false questions are worth 2 points each, three-choice multiple choice questions are worth 3 points each, five-choice multiple choice questions are worth 6 points each. The maximum possible score is 128. When the exam was given, the mean was 93.8; the median was 97. Click here to see the formula sheet that came with the exam.

(a) away from the plate (b) toward the plate (c) zero

The figure to the right shows the electric field lines for two charges. The arrows have been intentionally omitted.

Compare |Q_{A}|, the magnitude of charge A, with Q_{B} the magnitude of charge B.

(a) |Q_{A}| < |Q_{B}| (b) |Q_{A}| = |Q_{B}| (c) |Q_{A}| > |Q_{B}|

(T) True (F) False

(a) increase. (b) remain the same. (c) decrease.

Four point charges are arranged in this configuration. All four are equal in magnitude but a and c are negative while b and d are positive.

In which direction does the electric force on charge c point due to all other charges?

(a) up (b) down (c) zero

Compare V_{A} the electric potential at point A with V_{B} the electric potential at point B.

(a) V_{A} < V_{B} (b) V_{A} = V_{B} (c) V_{A} > V_{B}

(a) positive work. (b) zero work. (c) negative work.

Three charges are placed located as shown in the figure to the right. The grid spacing is in meters.

Calculate the x-component of the net force on the bottom charge at (0,-3) due to the two charges on the x-axis.

(a) F_{x} = -4.66 × 10^{-2} N (b) F_{x} = -3.50 × 10^{-2} N (c) F_{x} = 0 (d) F_{x} = +3.50 × 10^{-2} N (e) F_{x} = +4.66 × 10^{-2} N

(a) F_{x} = -4.66 × 10^{-2} N (b) F_{x} = -3.50 × 10^{-2} (c) F_{x} = 0 (d) F_{x} = +3.50 × 10^{-2} N (e) F_{x} = +4.66 × 10^{-2} N

(a) -4.05 × 10^{4} V (b) -1.35 × 10^{4} V (c) 0 (d) +1.35 × 10^{4} V (e) +4.05 × 10^{4} V

A capacitor array is shown in the figure to the right. Each capacitor is identical with capacitance C = 4µF. An unknown potential difference V is maintained by a power supply.

What is the equivalent capacitance of the network?

(a) C_{eq} = 2.18 µF (b) C_{eq} = 4.22 µF (c) C_{eq} = 5.91 µF (d) C_{eq} = 9.63 µF (e) C_{eq} = 25.6 µF

(a) 1.43 V (b) 4.22 V (c) 6.88 V (d) 9.63 V (e) 18.1 V

(a) increases. (b) decreases. (c) remains the same

(a) U_{tot} = 3.11 mJ (b) U_{tot} = 10.6 mJ (c) U_{tot} = 14.5 mJ

How much work did it require to assemble the charge distribution shown at the right?

(a) W = -67.2 nJ (b) W = -13.9 nJ (c) W = +48.8 nJ (d) W = +76.2 nJ (e) W = +89.3 nJ

(a) V(0,0) = -5.90 V (b) V(0,0) = -4.33 V (c) V(0,0) = +0.15 V (d) V(0,0) = +2.22 V (e) V(0,0) = +3.02 V

(a) (b) (c)

In this circuit, the switch has been open for a long time so that the capacitor is uncharged.

What is the current through the battery immediately after the switch is closed?

(a) I(0+) = 0 (b) I(0+) = V/(2R) (c) I(0+) = V/R

(a) I(¥) = 0 (b) I(¥) = V/(2R) (c) I(¥) = V/R

(a) Q(¥) = CV/2 (b) Q(¥) = CV (c) Q(¥) = 2CV

(a) t = 13.8 µs (b) t = 27.7 µs (c) t = 55.5 µs (d) t = 88.1 µs (e) t = 113.2 µs

Which one of the following is a valid Kirchhoff voltage equation for the dotted outside loop?

(a) -E_{1} + I_{1}R_{1} + E_{4} - I_{4}R_{4} = 0 (b) -E_{1} + I_{1}R_{1} - E_{4} + I_{4}R_{4} = 0 (c) -E_{1} + I_{1}R_{1} - E_{4} - I_{4}R_{4} = 0

(a) I_{1} + I_{2} + I_{3} + I_{4} = 0 (b) I_{1} - I_{2} + I_{3} + I_{4} = 0 (c) I_{1} + I_{2} - I_{3} - I_{4} = 0

(a) |V_{AB}| = 25 volts (b) |V_{AB}| = 18 volts (c) |V_{AB}| = 15 volts (d) |V_{AB}| = 10 volts (e) |V_{AB}| = 5 volts

Compare I_{2}, the current through resistor 2, with I_{4}, the current through resistor 4.

(a) I_{2} > I_{4} (b) I_{2} = I_{4} (c) I_{2} < I_{4}

(a) V_{1} > V_{5} (b) V_{1} = V_{5} (c) V_{1} < V_{5}

(a) P = 3.92 W (b) P = 5.32 W (c) P = 11.9 W (d) P = 19.5 W (e) P = 24.8 W

(a) I_{2} = 0.216 A (b) I_{2} = 0.388 A (c) I_{2} = 0.459 A (d) I_{2} = 0.521 A (e) I_{2} = 0.692 A

This figure should look familiar from your discussion section work! Simplify the system of capacitors and determine the total electrical energy stored.

(a) U_{tot} = 1.01 × 10^{-4} J (b) U_{tot} = 2.98 × 10^{-4} J (c) U_{tot} = 4.18 × 10^{-4} J (d) U_{tot} = 5.39 × 10^{-4} J (e) U_{tot} = 6.86 × 10^{-4} J

(a) 9.52 V (b) 11.8 V (c) 14.1 V