If the electric flux entering and leaving an enclosed surface respectively are $ϕ_{1}$ and $ϕ_{2}$ the electric charge inside the surface will be:
$(ϕ_{1} + ϕ_{2}) ϵ_{0}$
$(ϕ_{2} - ϕ_{1}) ϵ_{0}$
$\frac{ϕ_{1} + ϕ_{2}}{ϵ_{0}}$
$\frac{ϕ_{2} - ϕ_{1}}{ϵ_{0}}$

Question

If the electric flux entering and leaving an enclosed surface respectively are $ϕ_{1}$ and $ϕ_{2}$ the electric charge inside the surface will be:
$(ϕ_{1} + ϕ_{2}) ϵ_{0}$
$(ϕ_{2} - ϕ_{1}) ϵ_{0}$
$\frac{ϕ_{1} + ϕ_{2}}{ϵ_{0}}$
$\frac{ϕ_{2} - ϕ_{1}}{ϵ_{0}}$

Toppr Admin · Accepted Answer

Net electric flux -x3D5-x3D5-2-x2212-x3D5-1Applying Gauss-apos-s law- -x3D5-x222B-x2192-E-nds-Qenclosed-x3F5-0So- Qenclosed-x3D5-x3F5-0-x3D5-2-x2212-x3D5-1-x3F5-0

If the electric flux entering and leaving an enclosed surface respectively are ϕ1 and ϕ2 the electric charge inside the surface will be:(ϕ1+ϕ2)ϵ0(ϕ2−ϕ1)ϵ0ϕ1+ϕ2ϵ0ϕ2−ϕ1ϵ0

Net electric flux ,ϕ=ϕ2−ϕ1Applying Gauss's law, ϕ=∫→E.^nds=Qenclosedϵ0So, Qenclosed=ϕϵ0=(ϕ2−ϕ1)ϵ0

If the electric flux entering and leaving an enclosed surface respectively are $ϕ_{1}$ and $ϕ_{2}$ the electric charge inside the surface will be:
$(ϕ_{1} + ϕ_{2}) ϵ_{0}$
$(ϕ_{2} - ϕ_{1}) ϵ_{0}$
$\frac{ϕ_{1} + ϕ_{2}}{ϵ_{0}}$
$\frac{ϕ_{2} - ϕ_{1}}{ϵ_{0}}$

Net electric flux , $ϕ = ϕ_{2} - ϕ_{1}$
Applying Gauss's law, $ϕ = \int \to E .^n d s = \frac{Q_{e n c l o s e d}}{ϵ_{0}}$
So, $Q_{e n c l o s e d} = ϕ ϵ_{0} = (ϕ_{2} - ϕ_{1}) ϵ_{0}$