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borishaifa [10]
3 years ago
9

A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge

Q as shown in Example 19.5. The particle, confined to move along the x axis, is displaced a small distance x along the axis (x is much smaller than a) and released. Show that the particle oscillates in simple harmonic and calculate its frequency.
Physics
1 answer:
navik [9.2K]3 years ago
5 0

The solution would be like this for this specific problem:

<span>
F=−</span>k∗x∗<span>q∗</span>Q<span>/(</span>+)<span>F−≈</span><span><span><span>k∗x∗<span>q∗</span>Q</span><span>/R3</span></span>[(</span>1−<span><span>3/2</span><span><span>*x2</span><span>/R3</span></span>]
</span><span>F=−</span><span><span>k∗x∗<span>q∗/</span>Q</span><span>R<span>3
</span></span></span><span>F=</span><span>ma
</span>−<span><span><span><span>k∗<span>q∗</span>Q</span><span>/R3</span></span>*</span>x</span>=<span>ma
</span>−k∗x=m∗<span>a
a</span>==<span><span><span>ω2</span>x
</span>ω</span><span>=(</span>k/<span>m<span>)<span><span>1/</span><span>2

</span></span></span></span>ω<span>=(</span><span>kqQ</span>/<span><span>R3</span><span>)<span><span>1/</span>2

</span></span></span>

<span>I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

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