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

Question

If the electric flux entering and leaving an enclosed surface respectively is φ 1 and φ 2 the electric charge inside the surface will be (A)  (φ 2-φ 1)ε 0  (B)  (φ 1+φ 2)/ε 0  (C)  (φ 2-φ 1)/ε 0  (D)  (φ 1+φ 2)ε 0

Tardigrade Admin · Accepted Answer

[From Gauss law, ( textCharge textenclosed/ text0) = Flux leaving the surface] (q/ε 0)=φ 2-φ 1 ⇒ q=(φ 2-φ 1)ε 0

Q. If the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ the electric charge inside the surface will be

$(ϕ_{2} - ϕ_{1}) ε_{0}$

$(ϕ_{1} + ϕ_{2}) / ε_{0}$

$(ϕ_{2} - ϕ_{1}) / ε_{0}$

$(ϕ_{1} + ϕ_{2}) ε_{0}$

[From Gauss law, $\frac{Charge enclosed}{_{0}}$ = Flux leaving the surface] $\frac{q}{ε _{0}} = ϕ_{2} - ϕ_{1}$ $\Rightarrow$ $q = (ϕ_{2} - ϕ_{1}) ε_{0}$

Q. If the electric flux entering and leaving an enclosed surface respectively is ϕ1​ and ϕ2​ the electric charge inside the surface will be

(ϕ2​−ϕ1​)ε0​

(ϕ1​+ϕ2​)/ε0​

(ϕ2​−ϕ1​)/ε0​

(ϕ1​+ϕ2​)ε0​