# Chapter 26 - The Electric Field - Conceptual Questions: 5

The proton will experience the same magnitude of force at a distance of 2 cm from the wire.

#### Work Step by Step

The force on the electron is $F = E~q$ The strength of the electric field at a distance $r_1$ from a charged wire is $E_1 = \frac{\lambda_1}{2\pi ~\epsilon_0~r_1}$ Note that an electron and a proton have the same magnitude of charge. After the charge density is doubled, if a proton is to experience the same magnitude of force, the strength of the electric field $E_2$ must be the same as $E_1$. $E_2 = E_1$ $\frac{\lambda_2}{2\pi ~\epsilon_0~r_2} = \frac{\lambda_1}{2\pi ~\epsilon_0~r_1}$ $\frac{2~\lambda_1}{2\pi ~\epsilon_0~r_2} = \frac{\lambda_1}{2\pi ~\epsilon_0~r_1}$ $r_2 = 2~r_1$ $r_2 = (2)(1~cm)$ $r_2 = 2~cm$ The proton will experience the same magnitude of force at a distance of 2 cm from the wire.

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