JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 13)
Two electrons each are fixed at a distance '2d'. 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. Proton will execute simple harmonic motion having angular frequency : (m = mass of charged particle)
$${\left( {{{2{q^2}} \over {\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}$$
$${\left( {{{{q^2}} \over {2\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}$$
$${\left( {{{2\pi {\varepsilon _0}m{d^3}} \over {{q^2}}}} \right)^{{1 \over 2}}}$$
$${\left( {{{\pi {\varepsilon _0}m{d^3}} \over {2{q^2}}}} \right)^{{1 \over 2}}}$$
Explanation
The arrangement of charges is shown below
_24th_February_Evening_Shift_en_13_1.png)
As we know that,
Coulomb's force between two charges. i.e., q1 and q2,
$$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {({d^2} + {x^2})}}$$ ..... (i)
Here, $${q_1} = {q_2} = q$$
Force in SHM, $$F = m{\omega ^2}x$$ ...... (ii)
Since, in order to have SHM +q should move downwards and force responsible for this will be only
$$F' = F\sin \theta + F\sin \theta = 2F\sin \theta $$ ..... (iii)
Using Eqs. (ii) and (iii), we get
$$2F\sin \theta = m{\omega ^2}x$$
$$ \Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}\sin \theta = m{\omega ^2}x$$
$$ \Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}.{x \over {{{({d^2} + {x^2})}^{1/2}}}} = m{\omega ^2}x$$
$$ \Rightarrow \omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {{{({d^2} + {x^2})}^{3/2}}m}}} \right)^{1/2}}$$
As, $$x < < d$$
$$\therefore$$ $$\omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {m{d^3}}}} \right)^{1/2}}$$
As we know that,
Coulomb's force between two charges. i.e., q1 and q2,
$$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {({d^2} + {x^2})}}$$ ..... (i)
Here, $${q_1} = {q_2} = q$$
Force in SHM, $$F = m{\omega ^2}x$$ ...... (ii)
Since, in order to have SHM +q should move downwards and force responsible for this will be only
$$F' = F\sin \theta + F\sin \theta = 2F\sin \theta $$ ..... (iii)
Using Eqs. (ii) and (iii), we get
$$2F\sin \theta = m{\omega ^2}x$$
$$ \Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}\sin \theta = m{\omega ^2}x$$
$$ \Rightarrow {2 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {({d^2} + {x^2})}}.{x \over {{{({d^2} + {x^2})}^{1/2}}}} = m{\omega ^2}x$$
$$ \Rightarrow \omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {{{({d^2} + {x^2})}^{3/2}}m}}} \right)^{1/2}}$$
As, $$x < < d$$
$$\therefore$$ $$\omega = {\left( {{1 \over {2\pi {\varepsilon _0}}}{{{q^2}} \over {m{d^3}}}} \right)^{1/2}}$$
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