Two small spheres spaced 20.0 cm apart have equal charge. how many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57 * 10-21 n?

890 Coulomb's law is F = k * q1 * q2 / r^2 where F = force k = Coulomb's constant 8.99x10^9 N m^2/C^2 q1, q2 = signed charges r = distance between charges Since the charges are the same, let's simplify the equation, solve for q, then substitute the known values and calculate. F = k * q1 * q2 / r^2 F = k * q^2 / r^2 F*r^2 = k * q^2 F*r^2/k = q^2 sqrt (F*r^2/k) = q sqrt (4.57x10^-21 N * (0.2m) ^2 / (8.99x10^9 N m^2/C^2)) = q sqrt ((4.57x10^-21 N * 0.04m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt ((1.828x10^-22 N*m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt (2.03337x10^-32 C^2) = q 1.42596x10^-16 C = q So each sphere has to have an excess of 1.42596x10^-16 Coulombs of electrons. A coulomb is 6.24150934x10^18 electrons, so let's do the multiplication: 1.42596x10^-16 * 6.24150934x10^18 = 8.9001426584664x10^2 = 890 So each sphere has an extra 890 electrons.