Question

Two pith balls of mass 10mg each are suspended by two threads from the same support are charged identically. They move apart by 0.08m and threads make an angle 60 degree with each other. Find the charge on each pith ball ?

Answer 1

Answer :

Given :

r = 0.08 m

q₁ = q₂

So,

$F = \frac{kq^{2} }{r^{2} }$

$q^{2} = \frac{fr^{2} }{k}$

Where, K = 9.0 x 10⁹ N

l Cos 30° = 0.04 m

l = $\frac{0.04}{Cos 30}$ = 0.046188 m

=> $\frac{F*l}{cos 30}=W*sin30*l$   (#In place of Omega sign (ω) i used W.)

=> $F=W*cos30*sin30$

=> F = $4.33*10^{-3}N$

$q^{2} = \frac{fr^{2} }{k}$

$q^{2} = \frac{4.33*10^{-3}*0.08*0.08 }{9*10^{9}}$

q = 0.55 μC