A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very fa

Question

13 days ago

10

r from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

1 answer:

Softa [913] · Answer 1 · 2026-02-20T04:32:57+03:00

To address this query, we will utilize principles related to electric fields, linear charge density, and electrostatic forces.

The electric field is described as:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

In this context:

$\lambda$ = Linear charge density

$\epsilon_0$ = Permittivity of free space

r = Distance

The expression for linear charge density can be articulated as:

It is defined as

$\lambda = \frac{q}{L}$

By substituting,

$E = \frac{\frac{q}{L}}{2\pi \epsilon_0 r}$

$E = \frac{q}{2\pi \epsilon_0 rL}$

The initial and final electric force can be represented based on charge and electric field through

$F_i = E_i q$

$F_f = E_f q$

Upon substituting the electric field value, we obtain:

$F_i = (\frac{q}{2\pi \epsilon_0 rL})q = (\frac{q^2}{2\pi \epsilon_0 rL})$

If the length is reduced to one-third, then

$F_f = (\frac{q}{2\pi \epsilon_0 r(L/3)})q = (\frac{3q^2}{2\pi \epsilon_0 rL})$

The electric force ratio becomes

$\frac{F_f}{F_i} = \frac{(\frac{3q^2}{2\pi \epsilon_0 rL})}{(\frac{q^2}{2\pi \epsilon_0 rL})}$

$\frac{F_f}{F_i} = 3$

Hence, the determined ratio is 3.