Can you help with Resultant Force Question?

Question

3 charges are arranged in a triangle. A positive charge of 5 uC is placed at the origin. A - charge of 2 uC is placed 20 cm away in the 1st quadrant. A 3rd charge of 5 uC is placed in the 1st quadrant 8.75 cm away from the - charge. The angle from the original + charge to the - charge to the other + charge is 1.396 radians. What is the resultant force exerted on the third charge by the other two? So far, I have F13 as 12.83 N and F23 as 5.13N. I am trying to find the magnitude of F3. Am I on the right track?I would appreciate an expert's help on this one.

Answer 1

Let Point A (origin) have +5 uC charge (charge 1), and point B be having +5 uC (charge 3), and point C be having ( - 2 uC) (charge 2). find the magnitude of resultant force on +5 uC (charge 3) by other two changes. Given that Angle ACB = 1.395 radians (I hope you have given this right)
(- 2 uC)
C
b=20cm a=8.75 cm
A
origin (+5 uC) B
c =? (+5 uC)

in Triangle ABC: c^2=a^2+b^2 –ab Cos (C) = 415.35
or c= 20.38 cm c =0.2038 METER ……..(1)
also a = c Cos (B) + b Cos (C) so we get
Cos (B) = 0.2577 ……………..(2)

force on B by A = Fab =9*10^9*(5 *10^-6)^2 / (0.2038)^2 N
= 5.417 N
force on B by C=Fcb= 9*10^9* (5 *10^-6)(-2 *10^-6)/(0.0875)^2 N
= 11.755 N
These are magnitude of forces on charge at B. Please note the force Fab is a repulsive force on B and it will directed “FROM A TO B outwards” at point B (now a vector). Force Fcb is an attractive force on B, whereby B will pull charge C (smaller) with that force, and it will be directed “FROM C TO B outwards” at point B (now a vector). Thus, two vectors at B will form an equivalent angle CBA (make diagram). In other words these 2 forces will be making a resultant through Cos (B) angle. If R vector is their resultant then R = Net Fb = Fab + Fcb (vector notation), its magnitude is given by

R^2 = (Fab)^2 + (Fcb)^2 – 2 (Fab) (Fcb) * Cos (B)
= (5.417)^2 + (11.755)^2 – 2 (5.417) (11.755) * 0.2577 (from eq (2))
= 134.70
The magnitude of net force on Charge B is R = 11.60 N

Answer 2

What you need to do is leave the force in vector form. For example, F13=Fx13 + Fy13
F23=Fx23 + Fy23
F3=(Fx13+Fx23) xdirection (Fy13+Fy23)ydirection

The magnitude F3=((F3x)^2+ (F3y)^2)^1/2