If the electric flux entering and leaving a closed surface are 6×106 and 9×106 S.I. units respectively, then the charge inside the surface of permittivity of free space ε0 is

Question

If the electric flux entering and leaving a closed surface are 6×106 and 9×106 S.I. units respectively, then the charge inside the surface of permittivity of free space ε0 is (A) ε0 ×106 (B) - ε0 ×106 (C) - 2ε0 ×106 (D) 3ε0 ×106

Tardigrade Admin · Accepted Answer

Net flux = leaving flux-entering flux φ=9 × 106-6 × 106=3 × 106 Here, φ=(q/ε0) q =φ ε0=3 × 106 × ε q =3 ε0 × 106

Q. If the electric flux entering and leaving a closed surface are $6 \times 1 0^{6}$ and $9 \times 1 0^{6}$ $S . I$ . units respectively, then the charge inside the surface of permittivity of free space $ε_{0}$ is

$ε_{0} \times 1 0^{6}$

$- ε_{0} \times 1 0^{6}$

$- 2 ε_{0} \times 1 0^{6}$

$3 ε_{0} \times 1 0^{6}$

$2 ε_{0} \times 1 0^{6}$

Net flux = leaving flux-entering flux
$ϕ = 9 \times 1 0^{6} - 6 \times 1 0^{6} = 3 \times 1 0^{6}$
Here, $ϕ = \frac{q}{ε _{0}}$
$q = ϕ ε_{0} = 3 \times 1 0^{6} \times ε$
$q = 3 ε_{0} \times 1 0^{6}$

Q. If the electric flux entering and leaving a closed surface are 6×106 and 9×106 S.I. units respectively, then the charge inside the surface of permittivity of free space ε0​ is

ε0​×106

−ε0​×106

−2ε0​×106

3ε0​×106

2ε0​×106

Net flux = leaving flux-entering flux ϕ=9×106−6×106=3×106 Here, ϕ=ε0​q​ q=ϕε0​=3×106×ε q=3ε0​×106