Question

A particle of mass m and charge -q moves diametrically through a uniformly charged sphere of radius R with total charge Q. The angular frequency of the particle’s simple harmonic motion, if it’s amplitude < R, is given by:
A.) \[\sqrt{\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{qQ}{mR}}\]
B.) \[\sqrt{\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{qQ}{m{{R}^{2}}}}\]
C.) \[\sqrt{\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{qQ}{m{{R}^{3}}}}\]
D.) \[\sqrt{\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{m}{mQ}}\]

Answer 1

Hint: Use Gauss’s law to find the electric field and acceleration of an object in oscillation to find the frequency of a particle.

Complete step by step answer:

Consider a sphere of radius R with total charge Q.

As we know,
\[F=ma\]
\[a=\dfrac{F}{m}\]
According to the Lorentz’s law, \[F=qE\]
Where q is the charge and E is the electric field.
Here we are using a charge with negative charge. Hence,
\[F=-qE\]
So, acceleration of the charge is
\[a=\dfrac{-qE}{m}\]…………………….(1)

According to Gauss's law.

\[\oint{\vec{E}.\vec{d}s=\dfrac{q}{{{\varepsilon }_{0}}}}\]

Here we are considering a sphere of radius R. So, the surface area will be \[4\pi {{R}^{2}}\]

\[E.4\pi {{R}^{2}}=\dfrac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}\]

We are considering an arbitrary small sphere which included the charge -q. Since it is the only negative charge occupied in the whole sphere, we can equate their charge density. From this we can find the charge enclosed.

Here, \[q=\dfrac{Q}{\dfrac{4}{3}\pi {{R}^{3}}}.\dfrac{4}{3}\pi {{r}^{3}}\]
\[E.4\pi {{r}^{2}}=\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{Q}{\dfrac{4}{3}\pi {{R}^{3}}}.\dfrac{4}{3}\pi {{r}^{3}}\]
Reduce this equation
\[E=\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{Qr}{4\pi {{R}^{3}}}\]…………………..(2)
We can assign equation 2 into equation 1. Therefore,
\[a=-\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{Qr}{4\pi {{R}^{3}}}\dfrac{q}{m}\]………………..(3)

We are considering that amplitude is smaller compared to the radius of the charged sphere. The maximum displacement is known as the amplitude. Here we can see the acceleration is directly proportional to the displacement and it is directed towards the mean position. These types of motion are known as simple harmonic motion. Hence, the charged particle will undergo simple harmonic motion.

Acceleration of any object at any point in its oscillation is given below,
\[a=-{{\omega }^{2}}x\], where \[\omega \] is the angular frequency, x is the displacement from the central position. We have been used r instead of x. So, the frequency will be,
\[\omega =\sqrt{-\dfrac{a}{r}}\]

We can calculate the frequency from equation 3.
\[\omega =\sqrt{\dfrac{Qq}{{{\varepsilon }_{0}}4\pi {{R}^{3}}m}}\]

Hence the option (C) is correct.

Note: Don’t forget to put the amplitude as r, since in the question itself shown that the amplitude of the particle is less than the radius of the sphere. If we are giving R instead of r, then the final answer will change accordingly and the solution will be wrong. Try to learn Gauss law and Lorentz law. We will use these laws extensively to find the electric field and potential of any objects.