JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 23)
A uniformly charged ring of radius 3a and total
charge q is placed in xy-plane centred at origin.
A point charge q is moving towards the ring
along the z-axis and has speed u at z = 4a. The
minimum value of u such that it crosses the
origin is :
$$\sqrt {{2 \over m}} {\left( {{2 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
$$\sqrt {{2 \over m}} {\left( {{1 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
$$\sqrt {{2 \over m}} {\left( {{1 \over {5}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
$$\sqrt {{2 \over m}} {\left( {{4 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
Explanation
Ui + Ki = Uf + Kf
$${{k{q^2}} \over {\sqrt {16{a^2} + 9{a^2}} }} + {1 \over 2}m{v^2} = {{k{q^2}} \over {3a}}$$
$${1 \over 2}m{v^2} = {{k{q^2}} \over a}\left( {{1 \over 3} - {1 \over 5}} \right) = {{2k{q^2}} \over {15a}}$$
$$v = \sqrt {{{4k{q^2}} \over {15ma}}} $$
$$ \therefore $$ v = $$\sqrt {{2 \over m}} {\left( {{{2{q^2}} \over {15 \times 4\pi {\varepsilon _0}a}}} \right)^{{1 \over 2}}}$$
$${{k{q^2}} \over {\sqrt {16{a^2} + 9{a^2}} }} + {1 \over 2}m{v^2} = {{k{q^2}} \over {3a}}$$
$${1 \over 2}m{v^2} = {{k{q^2}} \over a}\left( {{1 \over 3} - {1 \over 5}} \right) = {{2k{q^2}} \over {15a}}$$
$$v = \sqrt {{{4k{q^2}} \over {15ma}}} $$
$$ \therefore $$ v = $$\sqrt {{2 \over m}} {\left( {{{2{q^2}} \over {15 \times 4\pi {\varepsilon _0}a}}} \right)^{{1 \over 2}}}$$
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