A cube has sides of length L = 0.300 m . It is placed with one corner at the ori

Question

A cube has sides of length L = 0.300 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 =( -5.56 N/(Cm) )xi^+( 2.49 N/(Cm) )zk^.

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.

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Part B

Find the total electric charge inside the cube.

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A cube has sides of length L = 0.300 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 =( -5.56 N/(Cm) )xi^+( 2.49 N/(Cm) )zk^.

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)m2  

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Part B

Find the total electric charge inside the cube.

q = C  

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Explanation / Answer

(a) There is no field in the y direction, so the flux through the"y" faces is zero.

flux 1 and flux 3 are zero.

Then for side 6, which is at x = 0, the field through the face is zero, so the flux is also zero (because the field is propotinal to x).

Also for side 4, which is at z = 0, the field through the face is zero so the flux is also zero.

For the remaining two sides

side5: x = 0.300 so the field in the x direction is

E = -5.56 * 0.300

= -13.84 N/C.m   

area = 0.300 * 0.300 = 0.0900 m^2

flux = E * area = -1.67 *0.0900

= -0.150 N.m/C

For side 2: z = 0.300 so the field in the z direction is

E = 2.49 * 0.300 = 0.747 N/C.m

flux = 0.747 * 0.0900

= 0.067 N.m/C
So the answer to part a is

S1 = 0, S2 = 0.067, S3 = 0, S4 = 0, S5 = -0.150, S6 = 0.

(b) use Gauss law

charge inside = epsilon * total flux
  
= 8.85*10^-12 * (-0.150 + 0.067)

= -7.35*10^-13 C

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