I. E. Irodov Solution: 3.1

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: 3.1

Calculate the ratio of the electrostatic to gravitational interaction force between two electrons, between two protons. At what value of the specific charge \(\frac{q}{m}\) of a particle would these forces become equal (in their absolute values) in the case of interaction of identical particles?

Solution: 3.1

Here, \({m_e} = 9.11 \times {10^{ - 31}}kg\)
\({m_p} = 1.672 \times {10^{ - 27}}kg\)
\({q_e} = 1.602 \times {10^{ - 19}}C\)
\({q_p} = 1.602 \times {10^{ - 19}}C\)
\(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\)
\(G = 6.67 \times {10^{ - 11}}\)

Thus, for electron, \({F_e} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{q_e^2}}{{{r^2}}}\)
And for electron, \({F_g} = \frac{{Gm_e^2}}{{{r^2}}}\)

Therefore, for electron, \(\frac{{{F_e}}}{{{F_g}}} = \frac{{q_e^2}}{{\left( {4\pi {\varepsilon _0}} \right)Gm_e^2}}\)
\( = \frac{{{{\left( {1.602 \times {{10}^{ - 19}}C} \right)}^2}}}{{\left( {\frac{1}{{4\pi {\varepsilon _0}}}} \right) \times 6.67 \times {{10}^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}} \times {{\left( {9.11 \times {{10}^{ - 31}}kg} \right)}^2}}}\)
\( = 4 \times {10^{42}}\)

Similarly, for protons, \(\frac{{{F_e}}}{{{F_g}}} = \frac{{q_p^2}}{{\left( {4\pi {\varepsilon _0}} \right) \times G \times m_p^2}}\)
\( = \frac{{{{\left( {1.602 \times {{10}^{ - 19}}C} \right)}^2}}}{{\left( {\frac{1}{{9 \times {{10}^9}}}} \right) \times 6.67 \times {{10}^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}} \times {{\left( {1.672 \times {{10}^{ - 27}}kg} \right)}^2}}}\)
\( = 1 \times {10^{36}}\)

Now for, \({F_e} = {F_g}\)
or, \(\left( {\frac{1}{{4\pi {\varepsilon _0}}}} \right)\frac{{q_e^2}}{{{r^2}}} = \frac{{Gm_e^2}}{{{r^2}}}\)
or, \(\frac{{{q_e}}}{{{m_e}}} = \sqrt {4\pi {\varepsilon _0}G} \)
\( = \sqrt {\frac{{6.67 \times {{10}^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}}}{{\left( {9 \times {{10}^9}} \right)}}} \)
\( = 0.86 \times {10^{ - 10}}C.k{g^{ - 1}}\)