Consider three point charges located at the corners of a right triangle as shown
ID: 1767558 • Letter: C
Question
Consider three point charges located at the corners of a right triangle as shown in the figure, whereq1 = q3 = 8.00 ?C, q2 = ?2.00 ?C, and a = 0.180 m.
Find the resultant force exerted on q1
I need and explanation, I keep tyring it over and over but can't seem to understand
Consider three point charges located at the corners of a right triangle as shown in the figure, whereq1 = q3 = 8.00 ?C, q2 = ?2.00 ?C, and a = 0.180 m. Find the resultant force exerted on q1 I need and explanation, I keep tyring it over and over but can't seem to understandExplanation / Answer
force exerted by q2 on q1, F2 = k*q*q2/a^2
= 9*10^9*8*2*10^-12/0.18^2
F2 = 4.444 N (attractive force)
F2x = 0
F2y = 4.444 N
force exerted by q3 on q1, F3 = k*q*q3/2*a^2
= 9*10^9*8*8*10^-12/(2*0.18^2)
F3 = 8.889 N (repulsive force)
F3x = -F3*cos(45) = 8.889*cos(45) = -6.285 N
F3y = -F3*sin(45) = 8.889*sin(45) = -6.285 N
Fnetx = F2x + F3x = -6.285 N
Fnety = F2y + F3y = -1.841 N
Fnet = sqrt(Fnetx^2 + Fnety^2) = 6.55 N
direction, theta = tan^-1(Fnety/Fnetx) = 16.32 + 180 = 106.32 degrees with +x axis in counter clockwise