Asked by bossman
Two small pith balls, each of mass m = 11.2 g, are suspended from the ceiling of the physics lab by 1.7 m long fine strings and are not moving. If the angle which each string makes with the vertical is è = 30.4°, 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.)
Answers
Answered by
Elena
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•1.7•0.506) •sqrt(0.0112•9.8•0.577/9•10^9) =
=4.5•10^-6 C =4.5 μC.
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•1.7•0.506) •sqrt(0.0112•9.8•0.577/9•10^9) =
=4.5•10^-6 C =4.5 μC.