## A 1.3 cm � 1.3 cm � 1.3 cm box with its edges aligned with the xyz-axes is in the electric field E? =(360x+150)i^N/C, wh…

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A 1.3 cm � 1.3 cm � 1.3 cm box with its edges aligned with the xyz-axes is in the electric field E? =(360x+150)i^N/C, where x is in meters.

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2 weeks 2023-03-15T10:00:11+00:00 1 Answer 0 views Teacher 0

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Assuming the box is a cube, we can calculate the electric field at each of its eight corners by plugging in the coordinates into the expression for E? :

E? 1 = (360(0.013)i + 150) i^N/C = 4.68i + 150 i^N/C
E? 2 = (360(0.013)i) i^N/C = 4.68i i^N/C
E? 3 = (360(0.013)i + 150) j^N/C = 4.68j + 150 j^N/C
E? 4 = (360(0.013)i) j^N/C = 4.68j i^N/C
E? 5 = (360(0.013)i + 150) k^N/C = 4.68k + 150 k^N/C
E? 6 = (360(0.013)i) k^N/C = 4.68k i^N/C
E? 7 = (360(0.013)i + 150) i^N/C + (360(0.013)i) j^N/C = 4.68i + 4.68j + 150 i^N/C + 150 j^N/C
E? 8 = (360(0.013)i + 150) i^N/C + (360(0.013)i) k^N/C = 4.68i + 4.68k + 150 i^N/C + 150 k^N/C

The net electric field at each corner is the vector sum of the electric fields due to the box and the external electric field. Since the external electric field is only in the x-direction, we can simplify the expressions for the net electric fields at each corner:

E? net,1 = (4.68 + 360(0.013))i + 150 i^N/C = 9.48i + 150 i^N/C
E? net,2 = (4.68 – 360(0.013))i i^N/C = -3.72i i^N/C
E? net,3 = 4.68j + 150 j^N/C
E? net,4 = 4.68j i^N/C
E? net,5 = 4.68k + 150 k^N/C
E? net,6 = 4.68k i^N/C
E? net,7 = (4.68 + 360(0.013))i + 4.68j + 150 i^N/C + 150 j^N/C = 9.48i + 4.68j + 150 i^N/C + 150 j^N/C
E? net,8 = (4.68 + 360(0.013))i + 4.68k + 150 i^N/C + 150 k^N/C = 9.48i + 4.68k + 150 i^N/C + 150 k^N/C

The electric field is constant throughout the box, so the force on each charge inside the box is qE? net, where q is the charge of the particle. The force on each charge is in the same direction as the electric field, so the net force on the box is the sum of the forces on all the charges