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QJEEM19Phdc6c: JEE Main 2019 Physics - Class 12 JEEM19Phdc6c Solution: A uniformly charged ring of radius 3a and total charge q is…
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 :
Multiple Choice Options
A
2 m ( 2 15 q 2 4 π ε 0 a ) 1 / 2 \sqrt {{2 \over m}} {\left( {{2 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}} m 2 ( 15 2 4 π ε 0 a q 2 ) 1/2
B
2 m ( 1 15 q 2 4 π ε 0 a ) 1 / 2 \sqrt {{2 \over m}} {\left( {{1 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}} m 2 ( 15 1 4 π ε 0 a q 2 ) 1/2
C
2 m ( 1 5 q 2 4 π ε 0 a ) 1 / 2 \sqrt {{2 \over m}} {\left( {{1 \over {5}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}} m 2 ( 5 1 4 π ε 0 a q 2 ) 1/2
D
2 m ( 4 15 q 2 4 π ε 0 a ) 1 / 2 \sqrt {{2 \over m}} {\left( {{4 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}} m 2 ( 15 4 4 π ε 0 a q 2 ) 1/2
Step-by-Step Verified Solution
U
i + K
i = U
f + K
f k q 2 16 a 2 + 9 a 2 + 1 2 m v 2 = k q 2 3 a {{k{q^2}} \over {\sqrt {16{a^2} + 9{a^2}} }} + {1 \over 2}m{v^2} = {{k{q^2}} \over {3a}} 16 a 2 + 9 a 2 k q 2 + 2 1 m v 2 = 3 a k q 2 1 2 m v 2 = k q 2 a ( 1 3 − 1 5 ) = 2 k q 2 15 a {1 \over 2}m{v^2} = {{k{q^2}} \over a}\left( {{1 \over 3} - {1 \over 5}} \right) = {{2k{q^2}} \over {15a}} 2 1 m v 2 = a k q 2 ( 3 1 − 5 1 ) = 15 a 2 k q 2 v = 4 k q 2 15 m a v = \sqrt {{{4k{q^2}} \over {15ma}}} v = 15 ma 4 k q 2 ∴ \therefore ∴ v =
2 m ( 2 q 2 15 × 4 π ε 0 a ) 1 2 \sqrt {{2 \over m}} {\left( {{{2{q^2}} \over {15 \times 4\pi {\varepsilon _0}a}}} \right)^{{1 \over 2}}} m 2 ( 15 × 4 π ε 0 a 2 q 2 ) 2 1
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