Question

A charge of +1 coulomb is place at the 0 cm mark of a meter stick. A charge of +4 coulombs is placed at the 100 cm mark of the same meter stick. Where should a proton be placed on the meter stick so that the net force on it due to the two charges is 0? Answer in units of cm

Answer 1

As per the question the charge of one coulomb is at 0 cm of the metre stick.the second charge of 4 coulomb is situated at at 100 cm of metre stick.

hence the separation distance between them is 100 cm.

now as per the question a proton is set up between them in such a way that the net force on it is zero

let the charge of proton is q coulomb let the proton is situated at distance of x cm from the charge 1 coulomb.hence it is situated at a distance of 100-x cm from the charge 4 coulomb.

the force exerted by 1 coulomb on proton is-
`(1)/(4\pi\epsilon) (1*q)/(x^2)`

the force exerted by 4 coulomb on proton is-
`(1)/(4\pi\epsilon) (q*4)/([100-x]^2)`

as the net force is zero,hence-

`(1)/(4\pi\epsilon) (1*q)/(x^2)`
`=(1)/(4\pi\epsilon) (4*q)/([100-x]^2)`

`=(1)/(x^2) =(4)/([100-x]^2)`

`x^2=([100-x[^2)/(4)[/tex</p><p>[tex]x=(100-x)/(2)`

`3x=100`

`x=(100)/(3)`

`=33.33333`cm [ans]

Answer 2

`x = (100)/(3) cm`

Step-by-step explanation:

Let the proton is placed at the position x = x between two given charges

Now the force on this proton is zero due to two given charges

so we will have

`F_1 = F_2`

`(kq_1e)/(x^2) = (kq_2e)/((100 - x)^2)`

now plug in all data

`(1 C)/(x^2) = (4 C)/((100 - x)^2)`

square root both sides

`(1)/(x) = (2)/(100 - x)`

`100 - x = 2x`

`x = (100)/(3) cm`