I. E. Irodov Solution: 3.3

Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE,...

Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.

Problem: 3.3

Two small equally charged spheres, each of mass \(m\), are suspended from the same point by silk threads of length \(l\). The distance between the spheres \(x \ll l\). Find the rate \(\frac{{dq}}{{dt}}\) with which the charge leaks off each sphere if their approach velocity varies as \(v = \frac{a}{{\sqrt x }}\), where \(a\) is a constant. Solution: 3.3

Let the balls be deviated by an angle \(\theta \), from the vertical when separation between them equals to \(x\).
Applying Newton's second law of motion for any one of the sphere, we get,
\(T\cos \theta = mg\) ... ... ... (i)
\(T\sin \theta = {F_e}\) ... ... ... (ii)

From the equation (i) and (ii) we get,
\(\tan \theta = \frac{{{F_e}}}{{mg}}\) ... ... ... (iii)
But from figure we get, \(\tan \theta = \frac{x}{{2\sqrt {{l^2} - {{\left( {\frac{x}{2}} \right)}^2}} }} \simeq \frac{x}{{2l}}\) as \(x \ll l\) ... ... ... (iv)
From equation (iii) and (iv) we get
\(\frac{x}{{2l}} = \frac{{{F_e}}}{{mg}}\)
or, \({F_e} = \frac{{mgx}}{{2l}}\)
or, \(\frac{{{q^2}}}{{4\pi {\varepsilon _0}{x^2}}} = \frac{{mgx}}{{2l}}\)
Or, \({q^2} = \frac{{2\pi {\varepsilon _0}mg{x^3}}}{l}\) ... ... ... (v)
Differentiating equation (v) with respect to time
\(2q\frac{{dq}}{{dt}} = \frac{{2\pi {\varepsilon _0}mg}}{l} \times 3{x^2}\frac{{dx}}{{dt}}\)

According to the problem, \(\frac{{dx}}{{dt}} = v = \frac{a}{{\sqrt x }}\) as (approach velocity is \(\frac{{dx}}{{dt}}\))

Therefore, \(\sqrt {\frac{{2\pi {\varepsilon _0}mg}}{l}{x^3}} \frac{{dq}}{{dt}} = \frac{{3\pi {\varepsilon _0}mg}}{l}{x^2}\frac{a}{{\sqrt x }}\) Or, \(\frac{{dq}}{{dt}} = \frac{3}{2}a\sqrt {\frac{{2\pi {\varepsilon _0}mg}}{l}} \)