दो स्थिर इलेक्ट्रानों, जिनके बीच की दूरी '2d' है, के बीच इन्हें मिलाने वाली रेखा के मध्यबिन्दु पर तीसरा आवेश प्रोटॉन रखा है। इस प्रोटॉन को किसी लघु दूरी x (x << d) तक दोनों इलेक्ट्रॉनों को मिलाने वाली रेखा के लम्बवत् विस्थापित किया गया है। इसके कारण यह प्रोटॉन सरल आवर्त गति करने लगता है जिसकी कोणीय आवृत्ति होती है : (m = आवेशित कण की संहति)

CONCEPT:

• Coulomb law is defined as the force of the two-particle is directly proportional to the masses and inversely proportional to the square of the distance between the two particles and it is written as;

          \(F=\frac{1}{4\pi \epsilon_o}\frac{q_1q_2}{r^2}\)

         F is the force, q_1, and q₂ is the charge of mass 1 and mass 2 and r is the distance.

• Restoring force is written as;

          \(F_r = mgsin\theta\)

• Force has also been written as;

          F = mg

Here m is the mass and g is the acceleration due to gravity.

• Angular frequency\(\omega\)  is written as;

          \(\omega=\sqrt\frac{K}{m}\)

CALCULATION:

Given: Two electrons having charges q₁ and q₂ 

and 2d is the distance between the two charges.

A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges as shown in the figure,

Now, the restoring force is written as,

\(F_r = mgsin\theta\)

From the above figure we have;

\(F_r = 2Fsin\theta\)      ----(1)

Now, Using Columb's law we have;

\(F=\frac{1}{4\pi \epsilon_o}\frac{q_1q_2}{r^2}\)

⇒ \(F=\frac{1}{4\pi \epsilon_o}\frac{q^2}{x^2+d^2}\)     ----(2)

Now, On putting equation (2) in equation (1) we have;

\(F_r = 2(\frac{1}{4\pi \epsilon_o}\frac{q^2}{x^2+d^2})sin\theta\)     ----(3)

Here, \(sin\theta = \frac{x}{(d^2+x^2)^{1/2}}\)

Equation (3) becomes;

\(F_r = 2(\frac{1}{4\pi \epsilon_o}\frac{q^2}{x^2+d^2})\frac{x}{(d^2+x^2)^{1/2}}\)

⇒\(F_r = 2\frac{1}{4\pi \epsilon_o}\frac{q^2x}{(x^2+d^2)^{3/2}}\)

⇒ \(F_r = 2k\frac{q^2x}{(x^2+d^2)^{3/2}}\)

Here, \(k = \frac{1}{4\pi \epsilon_o}\)

when x<<d we have;

\(F_r = 2k\frac{q^2x}{d^{3}}\)

⇒ \(F_r =\frac{1}{2\pi \epsilon_o} \frac{q^2x}{d^{3}}\)      ----(4)

Now, as we know that F = Kx      ---(5)

From equation 4) and equation 5) we have;

⇒ \(Kx=\frac{1}{2\pi \epsilon_o} \frac{q^2x}{d^{3}}\)

⇒ \(K=\frac{1}{2\pi \epsilon_o} \frac{q^2}{d^{3}}\)

Now the angular frequency is written as;

\(\omega=\sqrt\frac{K}{m}\)

⇒ \(\omega=\sqrt\frac{​​\frac{1}{2\pi \epsilon_o} \frac{q^2}{d^{3}}}{m}\)

⇒ \(\omega= ​​{\left( {\frac{{{q^2}}}{{2{\rm{\pi }}{{\rm{\varepsilon }}_0}\,m{d^3}}}} \right)^{\frac{1}{2}}}\)

Hence, option 1) is the correct answer.