What is the electric flux linked with closed surface ?

Question

What is the electric flux linked with closed surface ? (A) 1011N m2C-1 (B) 1012N m2C-1 (C) 1010N m2C-1 (D) 8.86×1013 N m2 C-1

Tardigrade Admin · Accepted Answer

According to Gaussâs law Electric flux, φ=(q/ε0) where, q = total charge enclosed by closed surface ∴ φ=(1.25+7+1-0.4/ε0) =(8.85 C/8.85×10 -12 C2N-1m-2)=1012N m2 C-1

Q. What is the electric flux linked with closed surface ?

$1 0^{11} N m^{2} C^{- 1}$

$1 0^{12} N m^{2} C^{- 1}$

$1 0^{10} N m^{2} C^{- 1}$

$8.86 \times 1 0^{13} N m^{2} C^{- 1}$

According to Gauss’s law
Electric flux, $ϕ = \frac{q}{ε _{0}}$
where, q = total charge enclosed by closed surface
$∴ ϕ = \frac{1.25 + 7 + 1 - 0.4}{ε _{0}} < b r / >= \frac{8.85 C}{8.85 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2}} = 1 0^{12} N m^{2} C^{- 1}$

Q. What is the electric flux linked with closed surface ?

1011Nm2C−1

1012Nm2C−1

1010Nm2C−1

8.86×1013Nm2C−1

According to Gauss’s law Electric flux, ϕ=ε0​q​ where, q = total charge enclosed by closed surface ∴ϕ=ε0​1.25+7+1−0.4​<br/>=8.85×10−12C2N−1m−28.85C​=1012Nm2C−1

$1 0^{11} N m^{2} C^{- 1}$

$1 0^{12} N m^{2} C^{- 1}$

$1 0^{10} N m^{2} C^{- 1}$

$8.86 \times 1 0^{13} N m^{2} C^{- 1}$

According to Gauss’s law
Electric flux, $ϕ = \frac{q}{ε _{0}}$
where, q = total charge enclosed by closed surface
$∴ ϕ = \frac{1.25 + 7 + 1 - 0.4}{ε _{0}} < b r / >= \frac{8.85 C}{8.85 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2}} = 1 0^{12} N m^{2} C^{- 1}$