Instructions: Circle the letter corresponding to the correct answer choices. If
ID: 1512911 • Letter: I
Question
Instructions: Circle the letter corresponding to the correct answer choices. If you are sure that answer is correct and does not match any of those given, then write your answer on the blank space in f. For problems were you need to write in answers, correct units (Si system) must accompany the units; Leaving off the units will cost you 1/3 rd the points that answer, if the number is correct. Two negative, equally charged drops water, 3.80 mm apart, exert an electrostatic force of 2.8 Times 10^-9 N on each other. Determine the net charge on each drop. 4.34 Times 10^(-2) nC. 6.7 Times 10^(-12)C. 7.69 Times 10^(3) pC. 7.17 Times 10^(-3) pC 8.19 Times 10^(3) pC. Then determine the number of electron that had been added to each drop to cause the net charge on it. 1.28 Times 10^(8) 3.73 Times 10^(9). 2.26 Times 10^(3). 9.26 Times 10^(6) 4.22 Times 10^2(7). A thin straight glass rod. 920 cm long, is placed on the x-axis so that it is symmetric about the y axis. An electrical charge Q = - 478 mC, is uniformly distributed over it. Determine the charge density of the rod. -5.20 Times 10^(-2)C/m. 1.20 Times 10^(-3)C. -6.20 Times 10^(-2)C/m. -4.20 Times 10^(-3)C/m. 1.20 Times 10^(-3)C. Determine the electric field at a point p located at (0.00, 1.50 cm). 9.28 Times 10^(-8)C. 6.72 Times 10^(9)N/C. -6.23 Times 10^(10)N/C. 2.04 Times 10^(7)N/C. Assignment 4 represent a charged (Q = -7.26 mC) conducting solid rod coaxial with a charged (q = + 7 26 mC) cylindrical conducting shell The radius of the cylindrical rod is 9.40 cm and the radius of the conducting cylindrical shell is 26.4 cm. Determine the electric field at a distance of ISO cm, from the common axis Tlie length of each object is 35.0 m 3.04 Times 10^(7) N/C. -4.04 Times 10^(8)N/C. 4.77 Times 10^(9) N/C. -l 04 Times 10^(7) N/C 9 04Times 10^(6)N/C. Determine the direction of the electric field at r = 5.20 cm from the common axis. 8.26 C/N 5.19 C. 3.38 N/C. 2.38C. 5.19C. Determine the magnitude of the electric field at r = 155.0 cm from the common axis. 8.26 C/N 5.19 C. 3.38 N/C. 2.38C 5.19C Suppose that following the CME it is found that at an altitude of 2500 m the electric field has magnitude 1100 N/C and at an altitude of 1500 m the magnitude is 3700 N/C. In both cases the direction of the electric field is vertically toward the earth's surface. Find the net amount of charge (also indicate the sign of the charge) contained in a cube with horizontal forces at altitudes of 2500 and 1500 m. Assume that the surface of the earth is flat and horizontal.3.34 Times 10^(-2) C. -5.76 Times 10^(-2) C. 7.69 Times 10^(3) pC. 7.47 Times 10^-3) pC -5.19 Times 10^(3) pC. The values of C_1, C_2, C_3, C_4 and C_5 are 250, 450, 330, 570, and 190 microfarad =, respectively. Determine the equivalent capacitance of c_2, C_4 and C_5. 9.15 Times 10^(-4) F.4.55 Times 10^(-7) F. 4.67 Times 10^(-6) F. 5,15 Times 10^(-4) F. 5.93 Times 10^(-4) F. Determine the equivalent capacitance C_A - B between terminals A and B. 9.93 Times 10^(-4) nF. 4.55 Times 10^(-7) F. 1.15 Times 10^(-4) F. 5.15 Times 10^(-4) F 7.93 Times 10^(-4) F. A rod in the shape of rectangular solid has cross-sectional area of 17.8 cm^2, length of 5.20 cm resistivity of 25.8 Times 10^-8 m. The material from which the block is made has 5.95 Times 10^28 conduction electrons/m^3. A potential differences of 52.6 m V is maintained between its ends. What is the current in the block? 6.98 Times 10^(3) A. 4.65 Times 10^(-1) A 5.81 Times 10^(-1) ma. 6.71 Times 10^(-30 mA. 9.31 times 10^(-2) A. Assume that the current density in the rod is uniform. What is its value? 6.761 Times 10^(-2) A. 4.65 Times 10^(-3)A. 5.81 Times 10^(-1) A.m^2. 6.71 Times 10^(-6) a/M^2. 3.92 Times 10^6a/m^2.Explanation / Answer
1)
Fe = k*q1*q2/r^2
q1 = q2 = q
Fe = 28.5*10^-9 N
r = 3.8*10^-3 m
28.5*10^-9 = 9*10^9*q^2/(3.8*10^-3)^2
q = 6.76*10^-12 C <<<----------answer
Q = ne
n = Q/n = 4.22*10^7 <<<<<<<<----------answer
++++++++++++++++++++
2)
charge density = charge / length
charge density = -478*10^-3/9.2 = -5.2*10^-2 Cm
option (a)
electric field E = lambda/(2*pi*eo*r)
E = -5.2*10^-2/(2*pi*8.85*10^-12*0.015)
E = -6.23*10^10 N/C
option (d)
+++++++++++++++++++
4)
length of the cube l = 2500-1500 = 1000m
electric flux = A*(E2-E1) = 1000^2*(3700-1100) = 2600000000 Nm^2/C
from gauss law
flux = Qin/eo
Qin = 2600000000*8.85*10^-12
Qin = -2.3*10^-2 C
++++++++++++++++++++
5)
C4 , C5 are in series
C45 = (C4*C5)/(C4+C5) = (570*190)/(570+190) = 142.5 micro farad
C245 = C2 + C45 = 450 + 142.5 = 5.93*10^-4 F
option(e)
______________
6)
Resistanc R = rho*L/A = 25.8*10^-8*0.052/(17.8*10^-4) = 7.54*10^-6
current I = V/R = 52.6*10^-3/( 7.54*10^-6)
I = 6.98*10^3 A
current density J = IA
J = 6.98*10^3/(17.8*10^-4)
J = 3.92*10^6 Am^2