Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A uniform thin ring of charge, with radius 4.50 cm and total charge 5.40 C, is l

ID: 1465534 • Letter: A

Question

A uniform thin ring of charge, with radius 4.50 cm and total charge 5.40 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 13.0 g and charge 6.00 C is on the negative x-axis, initially very far from the ring (effectively at .) 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

the total energy KE = PE

PE = kq1q2 / r and KE = 1/2mv2

1 / 2 m v2 = kq1q2 / r

1 / 2 X 0.013 X v2 = 9 X 109 X 5.4 X 10-6 X 6 X 10-6 / 0.045

6.5 X 10-3 X v2 = 291.6 X 10-3 / 0.045

6.5 X v2 = 291.6 / 0.045

v2 = 291.6 / ( 0.045 X 6.5 )

v2 = 291.6 / 0.2925

v2 = 996.92

v = 31.57 m /s

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote