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

Answer:

$x = \frac{100}{3} cm$

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

$F_1 = F_2$

$\frac{kq_1e}{x^2} = \frac{kq_2e}{(100 - x)^2}$

now plug in all data

$\frac{1 C}{x^2} = \frac{4 C}{(100 - x)^2}$

square root both sides

$\frac{1}{x} = \frac{2}{100 - x}$

$100 - x = 2x$

$x = \frac{100}{3} cm$

Answer 2

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-$\frac{1}{4\pi\epsilon} \frac{1*q}{x^2}$

the force exerted by 4 coulomb on proton is-$\frac{1}{4\pi\epsilon} \frac{q*4}{[100-x]^2}$

as the net force is zero,hence-

$\frac{1}{4\pi\epsilon} \frac{1*q}{x^2}$$=\frac{1}{4\pi\epsilon} \frac{4*q}{[100-x]^2}$

$=\frac{1}{x^2} =\frac{4}{[100-x]^2}$

$x^2=\frac{[100-x[^2}{4}[/tex[tex]x=\frac{100-x}{2}$

$3x=100$

$x=\frac{100}{3}$

$=33.33333$cm   [ans]