Question

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 far 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?

Answer 1

To solve this problem we will apply the concepts related to the electric field, linear charge density and electrostatic force.

The electric field is

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

Here,

$\lambda$= Linear charge density

$\epsilon_0$ = Permittivity of free space

r = Distance

The linear charge density can be written as,

Linear charge density is given as

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

Replacing,

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

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

The initial and final electric Force can be written as function of the charge and the electric field as

$F_i = E_i q$

$F_f = E_f q$

If we replace the value for the electric field we have,

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

Length is one third at the end, then

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

The ratio of the force is

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

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

Therefore the required ratio is 3