Physics 100B Exam - Answer Key

PHYSICS 100B EXAM I Spring NAME KEY Problem I Value Score 30 30 III 20 IV 20 TOTAL 100 change IV 0.79.7 1. Giving or receiving aid in any examination is cause for dismissal from the university. 2. Perform the necessary calculation in the spaces provided. If additional space is required, use the backs of the questions sheets. 3. All work must be shown in order to receive FULL credit. Work must be legible and comprehensible, and answers should be clearly indicated. USEFUL EQUATIONS AND CONSTANTS k9.0 X 10 Nm²/C2 qe-1.6 X 10-19C m.9.1 X 10 kg m, 1.7 X 10-27kg 1 eV 1.60 x 10-1 q+1.6 X 10-19C EnetE+E+E+.... F12=1192 E-E = VVVW/q, where W is the work done by the electric force to move the charge from b to a Was = qEd V=PE 7 Vou = 7 E = Vat d V = (point charge) KE = 2 C= C14 Ark d 2 2 2 C 11 Problem 1. (30 points) Two point charges q8.0 X 10°C and q+ 6.0 X 10 are placed 0.04 m apart on a straight line, as shown below. 9,=-8x10°C 92=+6x10°C E A 0.04 m 0.02m a) (8 points) Calculate the electric field E, at point A due to charge q₁. Point A is 0.02 m away from charge q2, as shown above. Specify the magnitude and the direction of the electric field. =11 2 =2×10 = 91 (r+d)2 = (9x10 Nm² C2 -6 8x10°C (0.06m) b) (8 points) Calculate the electric field E, at point A due to charge q₂. Specify the magnitude and the direction of the electric field. 92 d2 = = Ax 10 N (2 -2-6 6x 10 22 8 1.35x10 C c) (6 points) Calculate the net electric field E at point A. Specify both the magnitude and direction. Enet = -1 + = 2×10+13.5×10분 = +11.5×10= 1.15×10 C Enet d) (8 points) If an electron (qe-1.6 X 10-19C) is placed at point A, calculate the magnitude and direction of the net force acting on the electron. =9Ẻ F= -(1.6x10°C) 1.15×10° = - 1.84 x10" N C 3 Problem II. (30 points) A point charge q₁ +30 X 10-°C is placed 32 cm from a second point charge q2 = -30 X 10-°C, as shown below. 0.16m -6 9,=+30x10°C mid point 927-30×10°C -0.32m a) (8 points) Calculate the net electric potential V, at point a, which is midway the two point charges. Va=k+k=(9,+92)=0V since 99, b) (8 points) Calculate the net electric potential V, at point b, which is 10 cm closer to charge qa, as shown above. -6 Nm -6 30×10°C Vo= + = 9x10 m² 700x1030x10 = 9x10 2 (2 (0.10m) 1Km²) (30x10°C) [0-22m 0.10m = -1.47×10° V 1 c) (9 points) Calculate the work done by the electric field to move the charge qa+1 X 10-°C from point a to point b. Wab = 93 (Va-Vo) = (1x10°C) (OV-(-1.47×10)] = = 1.47 J d) (5 points) Calculate the change in kinetic energy of the charge qa as it moves from point a to point b. Δ(KE) = KE-KE = Wab = a 1.47J Problem III. (20 points) An electric field of 8.50 X 10° V/m is desired between two parallel plates each of area 35.0 cm² and separated by 2.45 mm of air. a) (7 points) Calculate the potential difference between the plates. E=V= Ed = 250x10) (2.45x10m) = 2.082 x 10 V b) (7 points) What is the charge on each plate? C == & C= → Q = VA = (2.082×10V) (35×10) 4rk d 4ㅠ (9×10) (2.45×10m) = 2.63×10°C While the battery is still connected to the capacitor, a sheet of plastic is inserted be- tween the plates. The dielectric constant of plastic is 4. c) (6 points) What is the charge on each plate in the presence of the dielectric? Co →Q=kQ = 4x (2.63×10℃) = 1.05×10c -