Answer
The electric field strength 5.0 cm from the wire is 4000 N/C
Work Step by Step
The strength of the electric field at a distance $r$ from a charged wire is $E = \frac{\lambda}{2\pi ~\epsilon_0~r}$
Let $r_1 = 10.0~cm$. We can write an expression for the electric field strength at a distance of $r_1$:
$E_1 = \frac{\lambda}{2\pi ~\epsilon_0~r_1}$
Let $r_2 = 5.0~cm$. We can write an expression for the electric field strength at a distance of $r_2$:
$E_2 = \frac{\lambda}{2\pi ~\epsilon_0~r_2}$
We can divide $E_1$ by $E_2$:
$\frac{E_1}{E_2} = \frac{\frac{\lambda}{2\pi ~\epsilon_0~r_1}}{\frac{\lambda}{2\pi ~\epsilon_0~r_2}}$
$E_2 = \frac{r_1~E_1}{r_2}$
$E_2 = \frac{(10.0~cm)(2000~N/C)}{5.0~cm}$
$E_2 = 4000~N/C$
The electric field strength 5.0 cm from the wire is 4000 N/C.
