Question

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.

Answer 1

can you find the electric field strength along the x axis?

Answer 2

force on particle when it is displaced by a small dist x along the axis is

Answer 3

\[F=-kqQ/(X ^{2} + R ^{2})^{3/2} \]

Answer 4

since x is very small therefore neglecting it wrt R and we get \[F=-KqQx/R ^{3} = Ma\]

Answer 5

since a is propotional to -x therefore it is SHM
time period = 2pi/omega where omega = sq root(a/x)

Answer 6

Our colleague responded but has some observations
Note that the electric field produced by a ring uniformly loaded along its axis is given by:
\[E=k*x*q/(x^{2}+R^{2})^{3/2} \]
R is the ring radius
Thus, the electrical force that will act on the charge Q is
F = E * Q (along the radial direction as well)
\[F=-k*x*q*Q/(x^{2}+R^{2})^{3/2} \] (minus sign because it points in the opposite direction to the displacement)
When x << R, we can expand the denominator in series:
\[F=-k*x*q*Q/(x^{2}+R^{2})^{3/2} \]
\[F -\approx\frac{ k*x*q*Q }{ R^{3} } [(1-\frac{3}{2}\frac{x^{2}}{R^{3}} ]\]
Then
\[F =-\frac{ k*x*q*Q }{ R^{3} }\]
Now this is the resultant force acting on the charge Q, the second law of Newton,
\[F=ma\]
\[-\frac{k*q*Q}{ R^{3} }x =ma\]
In a simple harmonic oscillator (mass m and elastic constant k), the second law of Newton in provides:
\[-k*x=m*a\]
\[a=\frac{d^{2}x}{d t^{2}}=\omega^{2}x\] with
\[\omega=(k/m)^{1/2}\]
Comparing equation osclador hamõnico with the equation obtained for the load under the action of the force identified the frequency of oscillation

\[\omega=(kqQ/R^{3})^{1/2}\]
Hope I helped