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

The correct option is C (ϕ_2 ϕ_1)ε _0 According Gauss's law, change in flux =Q_inϵ_0 or (ϕ_2 ϕ_1)=Q_inϵ_0 or Q_in=(ϕ_2 ϕ_1)ϵ_0 ...

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Byju's Answer

Standard XII

Physics

Gauss's Law

lf the electr...

Question

lf the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ , the electric charge inside the surface will be :

(ϕ_{2} - ϕ_{1}) ε_{0}

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\frac{(ϕ_{2} + ϕ_{1})}{ε_{0}}

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\frac{(ϕ_{2} - ϕ_{1})}{ε_{0}}

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(ϕ_{2} + ϕ_{1}) ε_{0}

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Solution

The correct option is C $(ϕ_{2} - ϕ_{1}) ε_{0}$
According Gauss's law, change in flux $= \frac{Q_{i n}}{ϵ_{0}}$
or $(ϕ_{2} - ϕ_{1}) = \frac{Q_{i n}}{ϵ_{0}}$

or $Q_{i n} = (ϕ_{2} - ϕ_{1}) ϵ_{0}$

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lf the electric flux entering and leaving an enclosed surface respectively is ϕ1 and ϕ2, the electric charge inside the surface will be :

The correct option is C (ϕ2−ϕ1)ε0According Gauss's law, change in flux =Qinϵ0or (ϕ2−ϕ1)=Qinϵ0or Qin=(ϕ2−ϕ1)ϵ0

lf the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ , the electric charge inside the surface will be :

The correct option is C $(ϕ_{2} - ϕ_{1}) ε_{0}$
According Gauss's law, change in flux $= \frac{Q_{i n}}{ϵ_{0}}$
or $(ϕ_{2} - ϕ_{1}) = \frac{Q_{i n}}{ϵ_{0}}$

or $Q_{i n} = (ϕ_{2} - ϕ_{1}) ϵ_{0}$