The inward and outward electric flux for a closed surface in units of N-m2-c are respectively 8-103 and 4-103- Then the total charge inside the surface is -where - - permittivity constant-

The inward and outward electric flux from a closed surface are respectively $8 \times 10^{3}$ and $4 \times 10^{3}$ units. Then the net charge inside the closed surface is

Q. If the electric flux entering and leaving a closed surface are

6 \times 10^{6}

and

9 \times 10^{6} S I

units respectively, then the charge inside the surface of permittivity of free space

ω_{0}

Q. If the electric flux entering and leaving a closed surface are respectively of magnitude

ϕ_{1}

and

ϕ_{2}

then the electric charge inside the surface will be

Join BYJU'S Learning Program

The inward and outward electric flux for a closed surface in units of N−m2/c are respectively 8×103 and 4×103. Then the total charge inside the surface is [where ϵ = permittivity constant]

The correct option is D −4×103ϵ0 CBy Gauss's law ϕ=1ϵ0(Qenclosed) ⇒Qenclosed=ϕϵ0=(−8×103+4×103)ϵ0 =−4×103ϵ0 Coulomb.

The inward and outward electric flux for a closed surface in units of $N - m^{2} / c$ are respectively $8 \times 10^{3}$ and $4 \times 10^{3}$ . Then the total charge inside the surface is [where $ϵ$ = permittivity constant]

The correct option is D $- 4 \times 10^{3} ϵ_{0} C$
By Gauss's law $ϕ = \frac{1}{ϵ_{0}} (Q_{e n c l o s e d})$

$\Rightarrow Q_{e n c l o s e d} = ϕ ϵ_{0} = (- 8 \times 10^{3} + 4 \times 10^{3}) ϵ_{0}$
$= - 4 \times 10^{3} ϵ_{0}$ Coulomb.