Question

An electron experiences a force of magnitude F when it is$1cm$from a very long, charged wire with linear charge density$\lambda .$ If the charge density is doubled, at what distance from the wire will a proton experience a force of the same magnitudeF?

Short answer

The amount of distance from the wire when a proton experiences a force of the same magnitude F from the wire$=2\times {10}^{-2}\mathrm{m}$.

Step-by-step solution

Given information

For an electron: $r_1=1\mathrm{cm}=10-2\mathrm{m}$

Charge of electron$=-q$

Linear charge density in case of electron${\lambda }_1=\lambda$

Charge of proton$=+q$

Linear charge density${\lambda }_2=2\lambda$

Explanation

$$\begin{gathered} F=\frac{-e2\lambda }{4\pi e_0{10}^{-2}} \\ \Rightarrow F=\frac{e2\left(2\lambda \right)}{4\pi e_0{r}_2} \\ \Rightarrow \frac{-e2\lambda }{4\pi e_0{10}^{-2}}=\frac{e2\left(2\lambda \right)}{4\pi e_0{r}_2} \\ {r}_2=-2\times {10}^{-2}\mathrm{m} \\ \left|{\mathrm{r}}_2\right|=2\times {10}^{-2}\mathrm{m} \end{gathered}$$