If the charged ball is suspended be the string which is deflected by the angle α, the forces acting on it are: mg (downwards), tension T (along the string - to the pivot point), and F (electric force –along the line connecting the charges).
Projections on the horizontal and vertical axes are:
x: T•sin α = F, ….(1)
y: T•cosα = mg. ….(2)
Divide (1) by (2):
T•sin α/ T•cosα = F/mg,
tan α = F/mg.
Since
q1=q2=q.
r=2•L•sinα,
k=9•10^9 N•m²/C²
F =k•q1•q2/r² = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • sqrt(m•g•tanα/k)=
=(2•0.8•sin42.6) •sqrt(0.0177•9.8•tan42.6/9•10^9) =...
Pith Balls
Two small pith balls, each of mass m = 17.7 g, are suspended from the ceiling of the physics lab by 0.8 m long fine strings and are not moving. If the angle which each string makes with the vertical is è = 42.6°, and the charges on the two balls are equal, what is the magnitude of that charge (in µC)? (Please note that the unit of charge is micro-Coulomb here - some browsers might not display the Greek letter mu correctly and show it as an m.)
4 answers
sqrt((0.0177(2*0.8*sin42.6))(tan42.6)/9*10^9)
sqrt((0.0177(2*0.8*sin42.6))(tan42.6)/9*10^9)=1.5*10^-6
From what height can a 1-kg falling object cause fracture of the skull? Assume that the object is hard, that the area of contact with the skull is 1 cm2, and that the duration of impact is 10−3 sec.