The leakage current is :

\frac{Q}{K ρ \in_{0}}

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\frac{Q K}{ρ \in_{0}}

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\frac{Q}{ρ \in_{0}}

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\frac{Q \in_{0}}{ρ K}

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Solution

The correct option is A $\frac{Q}{K ρ \in_{0}}$
Leakage current, $I = \frac{V}{R}$
$R = \frac{ρ d}{A}$
$V = \frac{Q}{C}$ , $C = \frac{K A ϵ_{0}}{d}$
$V = \frac{Q d}{K A ϵ_{0}}$
Now leakage current $I = \frac{\frac{Q d}{K A ϵ_{0}}}{\frac{ρ d}{A}} = \frac{Q}{K ρ ϵ_{0}}$

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The leakage current is :

The correct option is A QKρ∈0Leakage current, I=VR R=ρdAV=QC, C=KAϵ0dV=QdKAϵ0Now leakage current I=QdKAϵ0ρdA=QKρϵ0

The correct option is A $\frac{Q}{K ρ \in_{0}}$
Leakage current, $I = \frac{V}{R}$
$R = \frac{ρ d}{A}$
$V = \frac{Q}{C}$ , $C = \frac{K A ϵ_{0}}{d}$
$V = \frac{Q d}{K A ϵ_{0}}$
Now leakage current $I = \frac{\frac{Q d}{K A ϵ_{0}}}{\frac{ρ d}{A}} = \frac{Q}{K ρ ϵ_{0}}$