An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.7 m and charge 0.027 mC. Suppose th

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An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.7 m and charge 0.027 mC. Suppose th

e electron is released from rest a distance 0.035 m from the ring center. It then oscillates through the ring center. Calculate its period. (The electron is always much closer to the ring center than a radius.)

Physics

1 answer:

DENIUS [597] · Answer 1 · 2023-05-30T10:09:24+03:00

The period is mathematically given as

t=7.34*10^{-8}s

<h3>What is the period?</h3>

Question Parameters:

Charge of radius 2.7 m and charge 0.027 mC.

The electron is released from rest at a distance of 0.035 m from the ring center.

Generally, the equation for the angular velocity is mathematically given as

$w^2=\frac{KQq}{R^3}$

Therefore

$W^2=\frac{9*10^9*9.1*10^{-5}*1.6*10^{-19}}{2.7^3*9.1*10^{-31}}$

w^2=7.316*10^{15}

w=8.55*10^2

Therefore

$2\pi/t=w\\\\2\pi/t=8.55*10^2$

t=7.34*10^{-8}s

In conclusion, The period is

t=7.34*10^{-8}s