Electric Field Calculation for Three Charges on a Horizontal Line

[jaejoon89]

"Three charges are placed at 100-m intervals along a horizontal line: a charge of –3.0 C
on the left, 2.0 C in the middle, and 1.0 C on the right. What is the electric field E on
the horizontal line halfway between the –3.0 C and 2.0 C charges?"



When I do the calculation E = E1 - E2 - E3 with E1 the electric field produced on the point from the -3.0 C charge, I get 3.2*10^6 N/C:

E = k[(3.0 C / (50 m)^2) - (2.0 C / (50 m)^2) - (1.0 C / (150 m)^2)] = 3.2*10^6 N/C.

However, the answer key says it is -1.8*10^7 N/C. How do you get this?

 

[Defennder]

E = k[(3.0 C / (50 m)^2) - (2.0 C / (50 m)^2) - (1.0 C / (150 m)^2)] = 3.2*10^6 N/C

This term should have a minus sign due to the value of the -3C charge.

 

[tiny-tim]

"Three charges are placed at 100-m intervals along a horizontal line: a charge of –3.0 C
on the left, 2.0 C in the middle, and 1.0 C on the right. What is the electric field E on
the horizontal line halfway between the –3.0 C and 2.0 C charges?"
…
When I do the calculation E = E1 - E2 - E3 with E1 the electric field produced on the point from the -3.0 C charge, I get 3.2*10^6 N/C:

E = k[(3.0 C / (50 m)^2) - (2.0 C / (50 m)^2) - (1.0 C / (150 m)^2)] = 3.2*10^6 N/C.

Hi jaejoon89!

Hint: if a test charge is positive, say, will each of the three charges move it to the left or the right?