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?

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1 Answer
Mar 7, 2018

Answer:

# 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"#