How would I answer this question about charges?

Question

Given $q_{1} = 2 q_{2} = 3 q_{3} = 4 n C$ $q_1=2q_2=3q_3=4nC$ , and each charge is $1 c m$ $1 cm$ away from the origin, what is the electric field magnitude at the origin?

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Answer 1 · 2018-03-07T00:58:33

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

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.

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

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

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

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

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