$2\mathrm{.}41\frac{N}{C}8\mathrm{.}85{10}^{-12}\frac{C^2}{N*m^2}=3\mathrm{.}141591\mathrm{.}47{10}^{-10}\frac{C}{m}.$

Part $A$

$At$ what distance from the wire $is$ the magnitude $of$ the electric field equal $to2\mathrm{.}41\frac{N}{C}?$

Use $8\mathrm{.}85{10}^{-12}\frac{C^2}{N*m^2}$ for the permittivity $of$ free space, and use $=3\mathrm{.}14159.$

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