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Physics-

General

Easy

Question

The electric flux for Gaussian surface A that enclose the charged particles in free space is (given q₁ = –14 nC, q₂ = 78.85 nC, q₃ = – 56 nC)

10³ Nm² C^–1
10³ CN^-1 m^–2
6.32 ' 10³ Nm² C^–1
6.32 ' 10³ CN^-1 m^–2

The correct answer is: 10³ Nm² C^–1

Flux is due to charges enclosed per $epsilon subscript 0 end subscript$
\ Total flux = $left parenthesis negative 14 plus 78.85 minus 56 right parenthesis n C divided by epsilon subscript 0 end subscript$
$equals 8.85 cross times 1 0 to the power of negative 9 end exponent C cross times fraction numerator 4 pi over denominator 4 pi epsilon subscript 0 end subscript end fraction equals 8.85 cross times 1 0 to the power of negative 9 end exponent cross times 9 cross times 1 0 to the power of 9 end exponent cross times 4 pi$
$equals 1000.4 N m to the power of 2 end exponent divided by C$ i..e. $1000 N m to the power of 2 end exponent C to the power of negative 1 end exponent$

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