If the electric flux entering and leaving an enclosed surface respectively are -1 and -2 the electric charge inside the surface will be-

The correct option is B (ϕ_2 ϕ_1)ϵ_0 Net electric flux , ϕ=ϕ_2 ϕ_1 Applying Gauss's law, ϕ=∫E⃗.n̂ds=Q_enclosedϵ_0 So, Q_enclosed=ϕϵ_0=(ϕ_2 ϕ ...

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Byju's Answer

Standard XII

Physics

Gauss's Law

If the electr...

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}

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(ϕ_{2} - ϕ_{1}) ϵ_{0}

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\frac{ϕ_{1} + ϕ_{2}}{ϵ_{0}}

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\frac{ϕ_{2} - ϕ_{1}}{ϵ_{0}}

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Solution

The correct option is B $(ϕ_{2} - ϕ_{1}) ϵ_{0}$
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, $Q_{e n c l o s e d} = ϕ ϵ_{0} = (ϕ_{2} - ϕ_{1}) ϵ_{0}$

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If the electric flux entering and leaving an enclosed surface respectively are ϕ1 and ϕ2 the electric charge inside the surface will be:

The correct option is B (ϕ2−ϕ1)ϵ0Net 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:

The correct option is B $(ϕ_{2} - ϕ_{1}) ϵ_{0}$
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, $Q_{e n c l o s e d} = ϕ ϵ_{0} = (ϕ_{2} - ϕ_{1}) ϵ_{0}$