Answer:
The distance of the proton is [tex]r_p =10 \ cm[/tex]
Explanation:
Generally the force experience by the electron is mathematically represented as
[tex]F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}[/tex]
Where [tex]\lambda _e[/tex] is the charge density of the charge wire before it is doubled
Also the force experience by the proton is mathematically represented as
[tex]F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}[/tex]
Given that the charge density is doubled i.e [tex]\lambda_p = 2 \lambda_e[/tex] and that the the force are equal then
[tex]\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}[/tex]
[tex]\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}[/tex]
[tex]r_p * \lambda_e =2 \lambda_e * r_e[/tex]
[tex]r_p =2 r_e[/tex]
Now given from the question that [tex]r_e[/tex] the distance of the electron from the charged wire is 5 cm
Then
[tex]r_p =2 (5)[/tex]
[tex]r_p =10 \ cm[/tex]
