Question

A +13 nC point charge is placed at the origin, and a +8 nC charge is placed on the x axis at x=5m. At what position on the x axis is the net electric field zero? (Be careful to keep track of the direction of the electric field of each particle.)

Answer 1

Step-by-step explanation:

Given

Charge
`q_1=13\ nC is placed at x=0`

Charge
`q_2=8\ nC is placed at x=5\ m`

Electric field because of both the charges will be away from them

Electric field because of charge
`q_1` at distance r from it

`E_1=(kq_1)/(r^2)`

`E_1=(9* 10^9* 13* 10^(-9))/(r^2)`

Electric Field due to charge
`q_2` at distance of 5-r from it

`E_2=(kq_2)/((5-r)^2)`

`E_2=(9* 10^9* 8* 10^(-9))/((5-r)^2)`

at this Point Net Electric field is zero i.e.

`E_1=E_2`

`(9* 10^9* 13* 10^(-9))/(r^2)=(9* 10^9* 8* 10^(-9))/((5-r)^2)`

`(5-r)/(r)=\sqrt{(8)/(13)}`

`5-r=0.784 r`

`5=1.784 r`

`r=(5)/(1.784)`

`r=2.80\ m`

Thus at
`x=2.8\ m` net electric field is zero