Find Magnitude of Force at Origin from 4 Charges

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To find the magnitude of the resulting force at the origin from the three charges excluding the one at (0, 0), the user calculated the forces using Coulomb's law, resulting in values for each charge. The user correctly identified the need to consider the angles of the forces to determine their vector directions. It was emphasized that the forces must be resolved into their x and y components before summing them to find the net force. The discussion highlighted the importance of vector addition, as simply adding magnitudes would not yield the correct result. Properly resolving the forces and their directions is crucial for an accurate calculation of the total force at the origin.
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Four charges are arranged in (x,y) plane:
-3nC at (-4, 1)
-4nC at (0, 0)
-6nC at (5, 3)
-7nC at (4, 5)

Find the magnitude of the resulting force charge at the origin(0, 0).

I know how to find radius:
R1(squared) = 17
R2(squared) = 0
R3(squared) = 34
R4(squared) = 41

But I don't know which charges to use in the equation to find four different forces:
F=(K*q1*q2)/r(sqared)
K = 8.99 * 10^9

Can someone please help me?
 
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Find the field at the origin by other charges exluding the one which is placed at the origin(* NOTE *) Add vectorially
 
posted the question wrong

Find the magnitude of the resulting force on the -4nC charge at the origin.

And radius' are suppose to be from that charge, so they are only 3 radius.
 
Yes , but F is vector u can not add only magnitudes it has to be done in proper manner
 
Tell me if I'm doing this right, I'm getting big numbers thta's why I have a doubt:

F1 = 8.99e9*4e-9*3e-9/17 = 6.346e-9
F2 = 8.99e9*4e-9*6e-9/34 = 6.346e-9
F3 = 8.99e9*4e-9*5e-9/41 = 4.385e-9

Ang1= tan^-1(1/-4) = -14.04
Ang2= tan^-1(3/5) = 30.96
Ang3= tan^-1(5/4) = 51.3

Am I right so far?
 
Let the forces be in form Kq1q2/r^2

Yes angles are correct but what are the directions of these vectors??



After finding the direction resolve the components along x&y axis And now add the components
 
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