# Resultant Force on Charge in the middle of square

by katia11
Tags: charge, force, middle, resultant, square
 P: 18 1. The problem statement, all variables and given/known data (a) Find the resultant force on a negative charge of -3q, placed at the center of a square of side b, which has charges on the 4 corners, respectively, q, 2q, -4q, 2q in clockwise order. (Magnitude and direction) 2. Relevant equations F= k*q1*q*(1/ r^2) Ftotal= F15 + F25 + F35 + F45 3. The attempt at a solution I drew the figure, and see that r = square root 2 *b* (1/2) (pythagorean theorem) One question is that F25 and F45 are equal and opposite. Do they cancel eachother out? If so, I wouldn't have to calculate the forces for those, yes? I'm just a little confused. Sorry if it's super easy.
HW Helper
P: 6,684
 Quote by katia11 1. The problem statement, all variables and given/known data (a) Find the resultant force on a negative charge of -3q, placed at the center of a square of side b, which has charges on the 4 corners, respectively, q, 2q, -4q, 2q in clockwise order. (Magnitude and direction) 2. Relevant equations F= k*q1*q*(1/ r^2) Ftotal= F15 + F25 + F35 + F45 3. The attempt at a solution I drew the figure, and see that r = square root 2 *b* (1/2)
I assume you mean:
$$r = \sqrt{2}b/2 = .707b$$
 One question is that F25 and F45 are equal and opposite. Do they cancel eachother out? If so, I wouldn't have to calculate the forces for those, yes?
Correct. Just draw each force vector and add them together (tail to head). You will see that the net force is in the direction of q.

AM
 P: 18 Ok awesome! So now I have F15= (2ke3q2)/(b2) N and F35=(2ke4q2)/(b2) N But how do I find the x and y components without numbers? Do I just put in F15x= F15 sin(135) F15y= F15 cos(135) and F35x= F35 sin(135) F35y= F35 cos(135) The fact that theres no value for q shouldn't be throwing me off this much, but it is.