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Answer :
Answer:
The electric flux through the surface is equal to 3.878 x 10³ Nm²/C
The field distance r is equal to half the length of each side of the cube.
From the area the length of each size was calculated and the field distance and charge were used in calculating the magnitude of the electric field vector which was found to be 202 x 10³ N/C
The total flux area available to this electric field is 6x32cm²
Explanation:
The full solution can be found in the attachment below.
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