a billion (1.0*10^9) electrons are removed from a neutral sphere and placed on a second neutral sphere 5cm away. What is the magnitude of the force the first exerts on the second? what is the magnitude of the force of the second sphere exerts on the first?
thanks!
2007-04-14 15:49:17 · 2 answers · asked by tmacfan1121 2 in Science & Mathematics ➔ Physics
the charge moved from one sphere to the other is
1E9 electrons*1.6E-19 C/electron = 1.6E-10 C
so the first sphere has charge 1.6E-10 C and the second has charge -1.6E-10 C
the electrostatic force felt by each is =
F= k * q1 * q2/r^2
r is distance between them = .05m
q2 and q1 are the same since you take absolute values of the charges. So
F = (9E9)(1.6E-10)^2/(0.05)^2
F= 9.216E-8 N
F = 921.6 micro Newtons
that is the force each exerts on the other since its an equal and opposite pairing. only difference is the direction (the opposite part), the one on the left exerts the force on the other to the right, and the other pushes the first to the left
2007-04-14 16:26:27 · answer #1 · answered by NArchy 3 · 0⤊ 0⤋
Don't you even read your physics book? One Coulomb is 6X10^18 electrons, so 10^9 electrons is about 1.66X10^(-10) coulombs. The force acting on each sphere (.05 m apart) is the same and is equal to
(9X10^9)*Q1*Q2/r² = 9.92X10^(-8) N
HTH
Doug
2007-04-14 23:26:13 · answer #2 · answered by doug_donaghue 7 · 0⤊ 1⤋