JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 17)
Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle '$$\theta$$' with the vertical?
$$x = {\left( {{{{q^2}l} \over {2\pi {\varepsilon _0}mg}}} \right)^{{1 \over 2}}}$$
$$x = {\left( {{{{q^2}l} \over {2\pi {\varepsilon _0}mg}}} \right)^{{1 \over 3}}}$$
$$x = {\left( {{{{q^2}{l^2}} \over {2\pi {\varepsilon _0}{m^2}g}}} \right)^{{1 \over 3}}}$$
$$x = {\left( {{{{q^2}{l^2}} \over {2\pi {\varepsilon _0}{m^2}{g^2}}}} \right)^{{1 \over 3}}}$$
Explanation
_27th_July_Morning_Shift_en_17_2.png)
$$T\cos \theta = mg$$
$$T\sin \theta = {{k{q^2}} \over {{x^2}}}$$
$$\tan \theta = {{k{q^2}} \over {{x^2}mg}}$$
$$\tan \theta \approx \sin \theta \approx {x \over {2L}}$$
$${x \over {2L}} = {{K{q^2}} \over {{x^2}mg}}$$
$$x = {\left( {{{{q^2}L} \over {2\pi {\varepsilon _0}mg}}} \right)^{1/3}}$$
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