A cube has sides of length L = 0.390 m . It is placed with one corner at the ori
ID: 1784611 • Letter: A
Question
A cube has sides of length L = 0.390 m . It is placed with one corner at the origin as shown in the figure (Figure 1) . The electric field is not uniform but is given by E =( -5.40 N/(Cm) )xi^+( 2.68 N/(Cm))zk^.
Figure 1 of 1
Part A
Find the electric flux through each of the six cube faces S1,S2,S3,S4,S5, and S6.
Enter your answers in ascending order separated by commas.
Part B
Find the total electric charge inside the cube.
A cube has sides of length L = 0.390 m . It is placed with one corner at the origin as shown in the figure (Figure 1) . The electric field is not uniform but is given by E =( -5.40 N/(Cm) )xi^+( 2.68 N/(Cm))zk^.
Figure 1 of 1
Part A
Find the electric flux through each of the six cube faces S1,S2,S3,S4,S5, and S6.
Enter your answers in ascending order separated by commas.
1,2,3,4,5,6= (N/C)m2Part B
Find the total electric charge inside the cube.
q = S2 (top) (back) S3 (right side) (left side) S4 (bottom) S5 (front)Explanation / Answer
A)
L = 0.390m
Let's call
-5.40 N/(C m) = a
2.68 N/(C m) = b
so
E = a x i + b z k
S1: only the x component, but x=0 everywhere on that surface: flux = Surface-integral = 0
S2: only the z-component "pierces"through there, and since z=L on that surface: Surface_integral = b L * L^2 = b L^3 = (2.68)(0.390)^3 = 0.15897
S3: same as S2 but with only x-component and x = L there =>
a L^3 = (-5.40)(0.390)^3 =- 0.3203
S4 (bottom) : only z-component of field pierces, but z=0 there, so : 0
S5 and S6 would need a y component of the field, but since that's zero: 0
Hence:
Ans: 0, 0.159, -0.320, 0, 0, 0 (N/C). m^2
B)
Adding these contributions gives: (a + b ) L^3, and according to Maxwell's first theorem in integral form:
Q = eps0 (a+b) L^3 = (8.854*10^-12) (-5.4 + 2.68) (0.390)^3 = -1.429*10^-12 C
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