Suppose that following the CME it is found that at an altitu
     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.  
 
  
  Solution
5.
C4 and C5 are in series ,so
1/C45 =1/570 + 1/190
C45=142.5 uF
C2 and C45 are in parallel
C245=142.5 + 410 =552.5 uF
C245 =5.53*10-4 F
b)
Answer is 1.15*10-4 F
C1 ,C245 and C3 are in parallel ,so equivalent capacitance
1/Cab =1/552.5 + 1/250 +1/390
Cab=119.4 uF =1.19*10-4 F
6.
Resistance
R=pL/A =(34.5*10-8)*0.052/(17.8*10-4)
R=1.007*10-5 ohms
Current in the block
I=E/R =52.6*10-3/1.007*10-5
I=5.22*103 A
b.
Current density
I=J/A =5.22*103/(17.8*10-4)
I=2.93*106A/m2


