Recent content by dancer2012

Ok so sin(θ)*string length=r SO.. F(Ty)=F(g)=mg F(Tx)=k(q^2/r^2) F(Tx)=F(Ty)*tan(θ) q=√((F(Tx)*r^2)/k) F(Tx)=mg*tan(θ) - - - - - - - - - F(Tx)=.012kg*9.81m/s^2=.11772N→A q=√((F(Tx)*(sin(θ)*L)^2)/k)----q=√(A*(sin(42.6)*1.2)^2/9E9)=2.938E-6C or 2.9μC Is 2.9 μC the correct answer or...

F(Tx)=mg*tan(θ) q=√((F(Tx)*r^2)/k) would r be mg/tan(θ)??

I understand that

Sorry here is the force diagram actually

The force diagram is attached F(Ty)=cos(θ)*F(T) - - - F(T)=F(Ty)/cos(θ) F(Tx)=sin(θ)*F(T) - - - F(T)=F(Tx)/sin(θ) So, F(Ty)/cos(θ)=F(Tx)/sin(θ) - - - F(Tx)=[F(Ty)*sin(θ)]/cos(θ) - - - F(Tx)=F(Ty)*tan(θ) Did I do all of that correctly?

Homework Statement Two small pith balls, each of mass m = 12 g, are suspended by 1.2 m fine (so that we can neglect their mass in this problem) strings and are not moving. If the angle that each string makes with the vertical is θ = 42.6°, and the charges on the two balls are equal, what is...