Asmall glass bead has been charged to + 30 nc . a tiny ball bearing 1.0 cm above the bead feels a 1.8×10^−2 n downward electric force. what is the charge on the ball bearing?

The charge on the ball bearing = 667 nC or 6.67 × 10⁻⁷ C

Explanation:

Coulombs law: The force of attraction or repulsion between two charges, Q₁ and Q₂, is directly proportional to the product of the charges and inversely proportional to the square of the distance between the two charges

From coulombs law, F =1/4πε₀ (Q₁Q₂/r²) Equation 1

making Q₂ the subject of the equation,

Q₂ =( F×r²)/(Q₁×1/4πε₀) Equation 2

Where F = force of attraction, Q₁ = charge of the bead, Q₂ = charge of the ball bearing, r = distance between the bead and the ball bearing.

Given: Q₁ = 30 nC = 30×10⁻⁹ C,  F = 1.8×10⁻² N,

             r = 1.0 cm = (1/100) = 0.01 m

Constant: 1/4πε₀ = 9×10⁹ Nm²/C²

Substituting these values intto equation 2,

Q₂ = [(1.8×10⁻²)(0.01)²]/[(9×10⁹)(30×10⁻⁹)]

Q₂ = (1.8×10⁻² × 0.0001)/270

Q₂ = 6.67 × 10⁻⁷ C. = 667 nC

Therefore the charge on the ball bearing = 667 nC

the part of the wave he is travelling on is the crest