Discussion
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 constant linear charge density λ, and an infinite plane with uniform surface charge density σ. If \(E_{1}(r_{0}) = E_{2}(r_{0}) = E_{3}(r_{0})\) at a given distance r0, then
Answer: C
The correct statement is \(E_{1}(r_{0}/2) = 2E_{2}(r_{0}/2)\), option (C).
The three fields are
\[ \begin{aligned} E_{1} &= \frac{Q}{4\pi\varepsilon_{0}r^{2}} \\ E_{2} &= \frac{\lambda}{2\pi\varepsilon_{0}r} \\ E_{3} &= \frac{\sigma}{2\varepsilon_{0}} \end{aligned} \]
Setting all three equal at \(r_{0}\) gives \(Q = 2\pi\sigma r_{0}^{2}\) and \(r_{0} = \lambda/(\pi\sigma)\), so (A), with \(4\sigma\pi r_{0}^{2}\), and (B), with \(2\pi\sigma\), are both off by a factor of 2.
Let the common value at \(r_{0}\) be E. Halving the distance multiplies \(E_{1}\) by 4 (inverse square) and \(E_{2}\) by 2 (inverse distance), while \(E_{3}\) does not change. So at \(r_{0}/2\): \(E_{1} = 4E\), \(E_{2} = 2E\), \(E_{3} = E\). Then \(E_{1} = 2E_{2}\), which is (C), while \(E_{2} = 2E_{3}\), not \(4E_{3}\), so (D) is false.
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