Question

Three charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q = 2.20

Answer 1

Answer 2

Here You have 3 forces, 3 sides of a triangle ( the distance L ),
since you have 2 find out force on charge (say Q1=7.00)

use coloumb's law F1=(q*Q1)/L :q=2.20uC, Q1=7.00uC =22 N
F2=(q*Q1)/L : same values = -40N
F=F1+F2

F=22-40

F=-18 N

To calculate the direction, use following formula

F=(q1*q2)/L*r^ : this r^ is the direction of the force exerted due to one point charge to the other point charge.

-18=(7*(-4))/0.7*r^2

r=1.4

Answer 3

Force	X component of the force	Y component of the force
Force on 7 $\mu$C by 2.2 $\mu$C	(9*109*7*10-6*2.2*10-6*cos 60) / (0.77)2	(9*109*7*10-6*2.2*10-6*sin 60) / (0.77)2
Force on 7 $\mu$C by -4 $\mu$C	(9*109*7*10-6*4*10-6*cos 120) / (0.77)2	- (9*109*7*10-6*4*10-6*sin 120) / (0.77)2

Force	X component of the force	Y component of the force
Force on 7 $\mu$C by 2.2 $\mu$C	0.1169 N	0.2025 N
Force on 7 $\mu$C by -4 $\mu$C	-0.2125 N	-0.3681 N
Total	?0.0956 N	?0.1656

Thus, total electric force on 7 $\mu$C is,

F = ((?0.0956)² + (?0.1656)²)^1/2

F = (0.018284)^1/2

F = 0.1352 N