Consider a uniform ring of charge lying in the x - y plane with radius a = 0.15
ID: 1736695 • Letter: C
Question
Consider a uniform ring of charge lying in thex-y plane with radius a = 0.15 mcentered at the origin. The zero of potential is at infinity. Thepotential V(O) at the center of the ring is 425V.
(a) What is the total charge Q on the ring?
Q = C
0 NO
(b) What is the potential at the point P located at(xP, yP,zP) = (0, 0, b), where b =0.31 m?
V(P) = V
0 NO
(c) A particle (q = +25 µC, m = 4.3 x10-4 kg) is released from rest from point P.What is the maximum speed this particle attains?
vmax = m/s
0 NO
Explanation / Answer
(a) Potential atthe center of the ring is given by V = k(Q/a2) Thencharge on the ring is Q = Va2/k Given V=425V Radiusof the ring is a = 0.15m k =9*109Nm2/C2 Q=(424V)(0.15m)2/(9*109Nm2/C2) = 1.06*10-9C (b) Potential ata point on the axis of the ring is given by V =k[Q/(a2+b2)] VP=(9*109Nm2/C2)[(1.06*10-9C)/{(0.15m)2+(0.31m)2}] =80.4V (c) This wecan find from the law of conservation of energy. At pointthe particle is at rest. So it will have only the potential energygiven by Vq When itwill have maximum velocity, its initial potential energy willtotally converts into kinetic energy(1/2)mvmax2 Vq =(1/2)mvmax2 vmax= [2Vq/m] =[(2)(80.4V)(25*10-6C)/(4.3*10-4kg)] = 3.05m/s Vq =(1/2)mvmax2 vmax= [2Vq/m] =[(2)(80.4V)(25*10-6C)/(4.3*10-4kg)] = 3.05m/s