By James Nearing,
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If m2 = M/2 this is nonsense. (c) This suffers from both of the difficulties of (a) and (b). b a c Electrostatics Example Still another example, but from electrostatics this time: Two thin circular rings have radii a and b and carry charges Q1 and Q2 distributed uniformly around them. The rings are positioned in two parallel planes a distance c apart and with axes coinciding. The problem is to compute the force of one ring on the other, and for the single non-zero component the answer is (perhaps) Fz = Q1 Q2 c 2π 2 0 π/2 0 dθ c2 + (b − a)2 + 4ab sin2 θ 3/2 .
34) to get Q1 Q2 Q1 Q2 c . 1 = 2 2 2π 0 2ac 4π 2 0 ac Now why should I believe this any more than I believed the original integral? When you are very close to one of the rings, it will look like a long, straight line charge and the linear charge density on it is then λ = Q1 /2πa. What is the electric field of an infinitely long uniform line charge? Er = λ/2π 0 r. So now at the distance c from this line charge you know the E-field and to get the force on Q2 you simply multiply this field by Q2 .
But there’s no friction or other such force to do this. OR If m1 = m2 , this is zero, but there are still unbalanced forces causing these masses to accelerate. (b) If the combination of masses is just right, for example m1 = 1 kg, m2 = 1 kg, and M = 2 kg, the denominator is zero. The expression for ax blows up — a very serious problem. OR If M is very large compared to the other masses, the denominator is negative, meaning that ax is negative and the acceleration is a braking. Without friction, this is impossible.