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Let $E_{1} (r), E_{2} (r) and E_{3} (r)$ be the respective electric fields at a distance $r$ from a point charge $Q$ , an infinitely long wire with uniform linear charge density $λ$ and an infinite plane with uniform charge density $σ . If E_{1} (r_{0}) = E_{2} (r_{0}) = E_{3} (r_{0})$ at some distance $r_{0},$ then

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$Q = 4 π r_{0}^{2} σ$

$λ = 2 π r_{0} σ$

$E_{1} (r_{0} / 2) = 2 E_{2} (r_{0} / 2)$

$E_{2} (r_{0} / 2) = 4 E_{3} (r_{0} / 2)$

answer is C.

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Detailed Solution

We have

$\begin{array}{l} \frac{Q}{4 π ε_{0} {r_{0}}^{2}} = \frac{λ}{2 π ε_{0} r_{o}} = \frac{σ}{2 ε_{0}} \\ \Rightarrow Q = 2 π r_{0}^{2} σ and λ = π r_{o} σ \end{array}$

Let $E_{1} (r_{o}) = E_{2} (r_{o}) = E_{3} (r_{o}) = E_{0} (say) .$ Then

$E_{1} (\frac{r_{o}}{2}) = 4 E_{0}, E_{2} (\frac{r_{o}}{2}) = 2 E_{0}, E_{3} (\frac{r_{o}}{2}) = E_{0}$

Thus we see that

$\frac{E_{1} (\frac{r_{o}}{2})}{E_{2} (\frac{r_{o}}{2})} = 2$

And $\frac{E_{2} (\frac{r_{o}}{2})}{E_{3} (\frac{r_{0}}{2})} = 2$

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