Question

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin. 2) What is the distance in meters?

Answer 1

The distance is 1.69 m.

Step-by-step explanation:

Given that,

First charge
`q_(1)= 3.8*10^(-6)\ C`

Second charge
`q_(2)=3.2*10^(-6)\ C`

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

`E_(1)=E_(2)`

`(kq_(1))/(x^2)=(kq_(2))/((d-x)^2)`

`(q_(1))/(q_(2))=((x)^2)/((d-x)^2)`

`\sqrt{(q_(1))/(q_(2))}=(x)/(d-x)`

`x=(d-x)*\sqrt{(q_(1))/(q_(2))}`

Put the value into the formula

`x=(3.25-x)*\sqrt{(3.8*10^(-6))/(3.2*10^(-6))}`

`x+x*\sqrt{(3.8*10^(-6))/(3.2*10^(-6))}=3.25*\sqrt{(3.8*10^(-6))/(3.2*10^(-6))}`

`x(1+\sqrt{(3.8*10^(-6))/(3.2*10^(-6))})=3.25*\sqrt{(3.8*10^(-6))/(3.2*10^(-6))}`

`x=\frac{3.25*\sqrt{(3.8*10^(-6))/(3.2*10^(-6))}}{(1+\sqrt{(3.8*10^(-6))/(3.2*10^(-6))})}`

`x=1.69\ m`

Hence, The distance is 1.69 m.