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Step-by-Step Solution
Step 1: Identify the System
A particle of mass 1 mg (i.e., 1 \times 10^{-6}\text{ kg} ) and charge q is placed midway between two identical charges q . The distance between the two fixed charges is 2 m, so the free particle is at a distance d = 1\,\text{m} from each charge. When the free particle is displaced slightly (by a small distance x ), it undergoes Simple Harmonic Motion (SHM).
Step 2: Write the Net Force
Let the displacement of the free particle from the midpoint be x (with x \ll 1\,\text{m} ). The net electrostatic force F on the free charged particle due to the two fixed charges can be written as:
F = \frac{k\,q^2}{(d + x)^2} \;-\; \frac{k\,q^2}{(d - x)^2},
where k = 9 \times 10^9\,\text{N\,m}^2\text{/C}^2 is the Coulomb’s constant.
Step 3: Approximate for Small Displacements
For small x , the resulting force can be expanded and simplified to a form proportional to -x , indicating SHM. A standard result (or by binomial approximation) leads to:
F = -\left(\frac{4\,k\,q^2}{d^3}\right)x.
This implies the acceleration a = \frac{F}{m} is:
a = -\left(\frac{4\,k\,q^2}{m\,d^3}\right)x.
Step 4: Identify the Angular Frequency
In SHM, a = -\omega^2\,x . Hence we compare and get:
\omega^2 = \frac{4\,k\,q^2}{m\,d^3} \quad\Longrightarrow\quad \omega = \sqrt{\frac{4\,k\,q^2}{m\,d^3}}.
Step 5: Substitute the Given Values
Given:
q^2 = 10\,\text{C}^2
m = 1 \times 10^{-6}\,\text{kg}
d = 1\,\text{m}
k = 9 \times 10^9\,\text{N\,m}^2\text{/C}^2
Substitute these into the formula for \omega :
\omega \;=\; \sqrt{\frac{4 \,\times\, 9\times10^9 \,\times\, 10}{(1\times10^{-6}) \,\times\, (1^3)}}
\;=\; \sqrt{ \frac{4 \,\times\, 9 \,\times\, 10 \,\times\, 10^9}{10^{-6}}}
\;=\; \sqrt{4 \,\times\, 9 \,\times\, 10 \,\times\, 10^{15}}.
Simplifying inside the square root:
= \sqrt{360 \,\times\, 10^{15}} \;=\; \sqrt{3.6 \,\times\, 10^{17}} \;=\; 6 \times 10^8\,\text{rad/s}.
Step 6: Express in the Form Requested
The angular frequency \omega = 6 \times 10^8 \,\text{rad/s} . We can write this as:
\omega \;=\; 6000 \times 10^5\, \text{rad/s}.
Hence, the answer in the form “_____ \times 10^5 rad/s” is 6000 \times 10^5 rad/s.
