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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 the 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.)

Answer :

Okpalawalter8go

The period is mathematically given as

t=7.34*10^{-8}s

What is the period?

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

[tex]w^2=\frac{KQq}{R^3}[/tex]

Therefore

[tex]W^2=\frac{9*10^9*9.1*10^{-5}*1.6*10^{-19}}{2.7^3*9.1*10^{-31}}[/tex]

w^2=7.316*10^{15}

w=8.55*10^2

Therefore

[tex]2\pi/t=w\\\\2\pi/t=8.55*10^2[/tex]

t=7.34*10^{-8}s

In conclusion, The period is

t=7.34*10^{-8}s

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