Asked by Rain

Three charges (q1 = 8.7 nC, q2 = -7.2 nC, and q3 = -4.8 nC) are at the corners of a square of side a = 9.2 cm, as shown in the picture. Determine the magnitude of the resultant electric force on each charge.
Answered by Elena
q1, q2, q3 –from the lower left corner – counterclockwise.
Origin of the coordinate system is at the corner where q1 is located.

F12=k•q1•q2/a² =
=9•10^9•8.7•10^-9•7.2•10^-9/9.2²•10^-4=
= 6.74•10^-5 N
F12x=0 F12y=6.74•10^-5 N.

F13=k•q1•q3/D(a√2)² =
=9•10^9•8.7•10^-9•4.8•10^-9/9.2²•10^-4•1.41=
=3.15•10^-5 N
F13x=F13y=3.15•10^-5•cos45=2.23•10^-5 N.

F1x= F12x+ F13x =0+2.23•10^-5 =2.23•10^-5 N.
F1y= F12y+ F13y= 6.74•10^-5 +2.23•10^-5 =8.97•10^-5 N.
F1=sqrt(F1x² + F1y²) =sqrt(2.23² +8.97²)•10^-5 =9.24•10^-5 N
The same procedure for other charges
Answered by Rain
Order, clockwise- q1, q2, q3....

Could you explain the procedure in words?
Answered by Rain
Never mind Elena, when I wrote it out on paper it made sense. Thank you very much! :)
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