Posted by physics on Monday, January 23, 2012 at 11:31am.
You need to specify the length of the sides of the triangle, R.
Then compute the two vector forces acting on the 2.2 uC charge. The 7.5 uC charge contributes a repulsive force
F1 = k*2.2*7.5*10^-12/R^2 Newtons
where k is the Coulomb constant and R is the side length in meters.
The other charge (-3.9 uC) will contribute an attractive force, given by the same formula,
F2 = k*3.9*2.2*10^-12/R^2 Newtons,
directed at 120 degrees to F1. Add them as vectors.
Related Questions
physics - Two point charges, 4.0×10-6 C and -1.0×10-6 C, are located...
physics - please check: Two point charges, initially 1 cm apart, are moved to a ...
gen. physics II - Two charges, +Q and ‑Q, are located two meters ...
physics - Three equal positive point charges of magnitude Q = 4.00ì C are...
physics - Three equal positive point charges of magnitude Q = 8.00ì C are...
Physics - Two charges, -2.0 micro C and -6.0 micro C, are located at (-0.65 m, 0...
Physics - Two point charges of magnitudes -6.25 nano coulomb and -12.5 nano ...
Physics - Three point charges, A = 2.2 µC, B = 7.5 µC, and C = -3.9...
Physics - Given a square of sidelength a = 4 cm. We place a charged particle at ...
science - Two point charges 8q and -2q are located at x=0 and x=l find the ...
For Further Reading