Consider a gold nucleus and an alpha particle. They exert an electrostatic repulsive force F on each other and are separated by distance r. The particle and the nucleus are being treated as point charges. when r = 5x10^-14m, F= 16N, and when r = 20x10^-14m, F = 1N. Use this information to calculate the values of the forces at distances r = 10x10^-14m and 15x10^-14m.
I'm really stuck on this one, can anyone help?
2007-12-31 01:53:34 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
at 4x distance then F is 16x smaller
therefore the force falls off as the inverse square of the distance.
so at r=10x10^-14m F=4N(double r=5x10^-14m so a quarter of the force)
and at r=15x10^-14m F=(16/9)N(3 times the radius so a ninth of the force)
2007-12-31 02:16:55 · answer #1 · answered by Clint 6 · 1⤊ 0⤋
when r1 = 05x10^-14m, F= 16N
(a) when r2 = 20x10^-14m, F = 1N
(b) when r3 = 10x10^-14m, F = ? N
(c) when r4 = 15x10^-14m, F = ? N
F changes as 1/r²,
(a) If r2 = 20 = 4r1, then F = 16/4² = 1 N
(b) If r3 = 10 = 2r1, then F = 16/2² = 4 N
(c) If r4 = 15 = 3r1, then F = 16/3² = 1.78 N
2007-12-31 13:15:35 · answer #2 · answered by MR.B 5 · 0⤊ 0⤋
The electrostatic force is
q1q2/( 4x (pi) x r^2 x E0)
Where E0 is a constant
Hence this is nice and easy.
2007-12-31 10:22:32 · answer #3 · answered by Mark G 7 · 0⤊ 0⤋
Electrostaitc force is inversely proportional to the distance between the charges square, i.e. force goes like 1/r^2. Since you know the forces at two radii, you cna use the 1/r^2 scaling to get the forces at the new radii.
Let F1 be force at radius r1 and Fn be teh unknown force at the new radius rn. Then
Fn/F1 = r1^2/rn^2 or Fn = F1*r1^2/rn^2
2007-12-31 10:09:49 · answer #4 · answered by nyphdinmd 7 · 0⤊ 0⤋
I dunno what the equation is, but I'm betting you don't need it. Use ratios:
F1/r1 = 16N/5x10^-14m and multiply that by one of the other radii, and do the same for F2/r2.
That might help.
2007-12-31 10:01:32 · answer #5 · answered by Bob S 2 · 0⤊ 0⤋