Question

An electron experiences a force of magnitude F when it is 5 cm 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 magnitude F?

Answer 1

Answer:

The  distance of the proton is  $r_p =10 \ cm$

Explanation:

Generally the force experience by the electron is mathematically represented as

         $F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}$

 Where  $\lambda _e$ is the charge density of the charge wire before it is doubled

         

Also the force experience by the proton is mathematically represented as

         $F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}$

Given that the charge density is doubled i.e $\lambda_p = 2 \lambda_e$ and that the the force are  equal then

      $\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}$

      $\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}$

      $r_p * \lambda_e =2 \lambda_e * r_e$

       $r_p =2 r_e$

Now given from the question that  $r_e$ the distance of the electron from the charged wire is  5 cm

 Then  

        $r_p =2 (5)$

         $r_p =10 \ cm$