The electric flux for Gaussian surface A that encloses the charged particles in free space is (Given: q1=-14 nC .q2=78.85 nC , q3=-56 nC )

Question

The electric flux for Gaussian surface A that encloses the charged particles in free space is (Given: q1=-14 nC .q2=78.85 nC , q3=-56 nC ) (A) 103 N m 2 C -1 (B) 103 C N -1 m -2 (C) 6.32 × 103 N m 2 C -1 (D) 6.32 × 103 C N -1 m -2

Tardigrade Admin · Accepted Answer

Flux is due to charges enclosed by the given surface. ∴ Total flux =(-14+78.85-56) nC / ε0 =8.85 × 10-9 C × (4 π/4 π ε0)=8.85 × 10-9 × 9 × 109 × 4 π =1000.4 Nm 2 / C i..e. 1000 Nm 2 C -1

Q. The electric flux for Gaussian surface $A$ that encloses the charged particles in free space is (Given: $q_{1} = - 14 n C$ $q_{2} = 78.85 n C, q_{3} = - 56 n C)$

$1 0^{3} N m^{2} C^{- 1}$

$1 0^{3} C N^{- 1} m^{- 2}$

$6.32 \times 1 0^{3} N m^{2} C^{- 1}$

$6.32 \times 1 0^{3} C N^{- 1} m^{- 2}$

Flux is due to charges enclosed by the given surface.
$∴$ Total flux $= (- 14 + 78.85 - 56) n C / ε_{0}$
$= 8.85 \times 1 0^{- 9} C \times \frac{4 π}{4 π ε _{0}} = 8.85 \times 1 0^{- 9} \times 9 \times 1 0^{9} \times 4 π$
$= 1000.4 N m^{2} / C$ i..e. $1000 N m^{2} C^{- 1}$

Q. The electric flux for Gaussian surface A that encloses the charged particles in free space is (Given: q1​=−14nC q2​=78.85nC,q3​=−56nC)

103Nm2C−1

103CN−1m−2

6.32×103Nm2C−1

6.32×103CN−1m−2