1 in = 2.54 cm, 1 ft = 12 in, 1 yd = 3 ft, 1 mi = 5280 ft, and 1.00 cal = 4.186
ID: 1522373 • Letter: 1
Question
1 in = 2.54 cm, 1 ft = 12 in, 1 yd = 3 ft, 1 mi = 5280 ft, and 1.00 cal = 4.186 J. The specific heat of water is 4.186 times 10^3 J/(kg-K), atmospheric pressure is 1.013 times 10^5 Pa, 1.0 liter = 1.0 times 10^-3 m^3, the speed of sound in air is 343 m/s at 20 degree C, and the threshold of hearing is 1.0 times 10^-12 W/m^2. The figure at right shows a configuration of three capacitors. The capacitors have capacitances C_1 = 6.0 muF, C_2 = 4.0 muF, and C_3 = 3.0 muF. What is the equivalent capacitance of this Cs combination of capacitors? A charge q_2 is placed at x = 0.24 m from q_1 which is at the origin. See the diagram below. A third charge. q_3 = -1.0 nC is placed at x = 0.16 m. The charge q_2 experiences no net electric force. What is q_1's charge? Answers scribbled on this page will not be graded.Explanation / Answer
1)
capacitance, C1=6*10^-6 F , C2=4*10^-6 F, C3=3*10^-6 F
C12=C1*C2/(C1+C2)
C12=(6*10^-6)*(4*10^-6)/(6*10^-6+4*10^-6)
c12=2.4*10^-6 F
and
C123=C12+C3
=2.4*10^-6 + 3*10^-6
=5.4*10^-6 F
=5.4uF
2)
electric force
F12=F23
k*q1*q2/r12^2 = k*q2*q3/r23^2
====>
q1/r12^2 = q3/r23^2
q1/(0.24)^2 = 1*10^-9/(0.08)^2
===> q1=9*10^-9 C
charge, q1=9*10^-9 C
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