Question

Two small pith balls, each of mass m = 14.2 g, are suspended from the ceiling of the physics lab by 0.5 m long fine strings and are not moving. If the angle which each string makes with the vertical is θθ = 29.1°, and the charges on the two balls are equal, what is the magnitude of that charge

Answer 1

Answer:

1.424 μC

Explanation:

I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),

tension T (acting along the string - to the pivot point), and

F (electric force – acting along the line connecting the charges).

We then have something like this

x: T•sin α = F,

y: T•cosα = mg.

Dividing the first one by the second one we have

T•sin α/ T•cosα = F/mg, ultimately,

tan α = F/mg.

Since we already know that

q1=q2=q, and

r=2•L•sinα,

k=9•10^9 N•m²/C²

Remember,

F =k•q1•q2/r², if we substitute for r, we have

F = k•q²/(2•L•sinα)².

tan α = F/mg =

= k•q²/(2•L•sinα)² •mg.

q = (2•L•sinα) • √(m•g•tanα/k)=

=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =

q = 0.486 • √(8.61•10^-12)

q = 0.486 • 2.93•10^-6

q = 1.424•10^-6 C

q = 1.424 μC.