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Two point charges of +5.00 μC and +8.00 μC are placed inside a cube of edge length...

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Arvy1 | Student | eNoter

Posted September 17, 2012 at 7:35 AM via web

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Two point charges of +5.00 μC and +8.00 μC are placed inside a cube of edge length 0.100 m. The net electric flux due to these charges is given by

A) 0.450 × 106 Nm2/C
B) 1.47 × 106 Nm2/C
C) 4.20 × 106 Nm2/C
D) 3.80 × 106 Nm2/C
E) 0.340 × 106 Nm2/C

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quantatanu | Student, Undergraduate | Valedictorian

Posted September 26, 2012 at 2:25 PM (Answer #1)

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There are two cases:

1) charges are well within.

2) charges are at corners.

for case 1)

Total Flux = (total charge)/(epsilon not)

               = [(5+8)*10^(-5)]/[8.85 * 10^(12)]

               =1.47 * 10^6 Nm^2/C

answer is (B)

for case 2)

If the charges are at corners then actually each cherge is shared by virtually 8 cubes, so you must divide the result by 8, so

Total Flux =(1/8) *  1.47 * 10^6 Nm^2/C

               = 1.83 * 10^6 Nm^2/C

as this result doesn't match any option so I think the charged are inside the cube in your problem. 

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So the case (1) is the actual case and the anser is (B)

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