A uniform thin ring of charge, with radius 5.90 cm and total charge 7.60 mu C, i

Question

A uniform thin ring of charge, with radius 5.90 cm and total charge 7.60 mu C, is located in the yz-plane and centered on the origin. (In other words, the x-axis passes through the center of the ring, and is perpendicular to the ring.) A particle of mass 11.1 g and charge 5.60 mu C is on the negative x-axis, initially very far from the ring (effectively at -infinity.) The particle is launched in the positive x direction with some initial velocity, toward the center of the ring. Calculate the minimum initial velocity required for the particle to get past the center of the ring and come out the other side.

Explanation / Answer

given that

R = 5.90 cm = 0.059 m

Qr = 7.60*10^(-6) C

mass of particle   m = 11.1 g = 0.011 kg

Qp = 5.60*10^(-6) C

according to energy conservation

initial KE = final PE

initial kinetic energy KE = 1/2*m*v^2

final potential energy PE = (k*Qr*Qp)/R

So

1/2*m*v^2 = (k*Qr*Qp)/R

1/2*0.011*v^2 = (8.85*10^(-12)* 7.60*10^(-6) *5.60*10^(-6)) / 0.059

v^2 = 6384*2*10^(-24)/0.011

v = sqrt(1160727.27*10^(-24))

v = 1077.37*10^(-12)

v = 1.07*10^(-9)

v = 1.07 nm

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