3) a small, 30 gram pith ball carrying 3 nc of charge is dangling from an insulating string. another charged pith ball carrying an unknown amount of charge is brought close to the first ball, causing it to shift away. when everything is in equilibrium, the two balls are horizontally separated by distance of 4 cm, and the string is tilted about 16 degrees away from the vertical direction. how much charge is on the second ball?

second bal charge is q₂ = 5 10⁻⁶ C

Explanation:

With this problem we must use Newton's second law and Coulomb's equation for electrostatic repulsion, the two balls have the same charge sign

         F = k q₁ q₂ / r²

Newton's second law, where the acceleration is zero, system in equilibrium

X axis

        F -Tx = 0

Axis y

       Ty -W = 0

We look for the components of stress with trigonometry

      sin θ = Tx / T

      Tx = T sin θ

      Cos θ = Ty / T

      Ty = T cos θ

We substitute and calculate

      F = T sin θ

      mg = T cos θ

      T = mg / cos θ

      F = mg sin θ / cos θ

      F = mg tan θ

Let's use Coulomb's law

       K q₁ q₂ / r² = mg tan θ

We have q₁ = 3 nC = 3 10⁻⁹ C,  calculate q2

       q₂ = mg tan θ (r² / k q₁)

       q₂ = mg r² tan θ / k q₁

       q₂ = 30 10⁻³ 9.8 (4 10⁻²)² tan 16 / (8.99 10⁹9 3 10⁻⁹)

       q₂ = 1.35 10⁻⁴ / 26.97

       q₂ = 5 10⁻⁶ C

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min