Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

q2 = 5.1μC

Explanation:

From a forces diagram we can stablish that:

Fe = T*sin(16)              (eq1)

T*cos(16) = m*g           (eq2)

Where Fe is the electric force and T is the tension of the string. Solving for Fe:

Fe = m*g*tan(16)

The magnitude of the electric force is calculated as:

$Fe = \frac{K*q1*q2}{d^2} = m*g*tan(16)$  Solving for q2:

$q2 = \frac{m*g*tan(16)*d^2}{K*q1}=5.1 \mu C$