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

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Electric Field Due to Charge Distributions - Approach

If the electr...

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

(ϕ_{1} + ϕ_{2}) / ε_{0}

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(ϕ_{1} - ϕ_{2}) / ε_{0}

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(ϕ_{1} + ϕ_{2}) ε_{0}

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(ϕ_{2} - ϕ_{1}) ε_{0}

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Solution

The correct option is C $(ϕ_{2} - ϕ_{1}) ε_{0}$
The net electric flux leaving the surface is $ϕ_{2} - ϕ_{1}$ .
Thus, by Gauss's law, $ϕ_{2} - ϕ_{1} = \frac{q}{ε_{o}} \Rightarrow q = (ϕ_{2} - ϕ_{1}) ε_{o}$ .

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If 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)ε0The net electric flux leaving the surface is ϕ2−ϕ1. Thus, by Gauss's law, ϕ2−ϕ1=qεo⇒q=(ϕ2−ϕ1)εo.

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

The correct option is C $(ϕ_{2} - ϕ_{1}) ε_{0}$
The net electric flux leaving the surface is $ϕ_{2} - ϕ_{1}$ .
Thus, by Gauss's law, $ϕ_{2} - ϕ_{1} = \frac{q}{ε_{o}} \Rightarrow q = (ϕ_{2} - ϕ_{1}) ε_{o}$ .