a)
Sorry for the design. I know I should refine a bit, but the time is much more important (for me) than such a detail.
b)
F13=kq1q3/r132
F13~3.54*10-9 N
c)
F23=kq2q3/r232
F23=16.2*10-9 N
d)
Ox: Rx=F13x=F13cos(alpha)~3.002*10-9 N
Oy: Ry=F23-F13y=F23-F13sin(alpha)~14.324*10-9 N
With:
cos(alpha)=0.848
sin(alpha)=0.53
R~14.635*10-9 N.
e)
The angle between R and Ox axis is: tan(beta)=Ry/Rx ---> beta~-78.16 deg.
If counterclockwise from Ox: ~101.84 deg.
Please, check the numerical calculation.
Show detailed solution k = 9 Times 10^9 Nm^2/C^2, e = plusminus 1.602 Times 10^-29 C,...
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