Question

Two engineering students, John with a weight of 92 kg and Mary with a weight of 46 kg, are 30 m apart. Suppose each has a 0.04% imbalance in the amount of positive and negative charge, one student being positive and the other negative. Estimate the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student. N

Answer 1

Answer:

F = 6.27 x 10 ¹⁹ N

Explanation:

Given

m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol

q₁ = % * [m * N * A * e / M ]  

q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₁ = 3.54 x 10⁶ C

q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₂ = 1.773 x 10⁶ C

Now to determine the electrostatic force con use the equation

F = K * q₁ * q₂ / d²

K = 8.99 x 10 ⁹

F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²

F = 6.27 x 10 ¹⁹ N