Question

How would I answer this question about charges?

Given `q_1=2q_2=3q_3=4nC`, and each charge is `1 cm` away from the origin, what is the electric field magnitude at the origin?

Answer 1

`E_o = (2397.33) \hatx \ "V/m" +(1798)\haty \ "V/m"`

Explanation:

Electric Field: `E = k(q)/(r^2)`

We simply sum the fields of the three charges at the origin.

Charge q1: `E_1 = k(q_1)/r^2`
Charge q2: `E_2 = k(q_2)/r^2`
Charge q3: `E_3 = k(q_3)/r^2`

The field at the origin: `E_(o) = E_1 + E_2 + E_3`
when `r_1=r_2=r_3=1` cm and `q_1=2q_2=3q_3=4` nC.

`E_o = k(((4xx10^(-9))/(1xx10^(-2))-(4/3xx10^(-9))/(1xx10^(-2)))\hatx+ (2xx10^(-9))/(1xx10^(-2))\haty)`

`E_o = (8.99xx10^(9)) (((4xx10^(-9))/(1xx10^(-2))-(4/3xx10^(-9))/(1xx10^(-2)))\hatx+ (2xx10^(-9))/(1xx10^(-2))\haty)`

`E_o = (8.99xx10^(9)) (((4xx10^(-7))-(4/3xx10^(-7)))\hatx+(2xx10^(-7))\haty)`

`E_o = (8.99xx10^(9)) ((8/3xx10^(-7))\hatx+(2xx10^(-7))\haty)`

`E_o = (2397.33)\hatx \ "V/m" + (1798)\haty \ "V/m"`