JEE Advance - Physics (2015 - Paper 1 Offline - No. 1)
The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density $$\lambda $$ are kept parallel to each other. In their resulting electric field, point charges $$q$$ and $$-q$$ are kept in equilibrium between them. The point charges are confined to move in the $$x$$ direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is (are)
Both charges execute simple harmonic motion
Both charges will continue moving in the direction of their displacement
Charge $$+q$$ executes simple harmonic motion while charge $$-q$$ continues moving in the direction of its displacement
Charge $$-q$$ executes simple harmonic motion while charge $$+q$$ continues moving in the direction of its displacement
Explanation
$$E = {\lambda \over {2\pi {\varepsilon _0}r}}$$
Case 1 : If q is shifted towards right by x, we get
$$F = {F_2} - {F_1} = {{\lambda q} \over {2\pi {\varepsilon _0}}}\left( {{1 \over {{d \over 2} - x}} - {1 \over {{d \over 2} + x}}} \right)$$ (towards left)
Case 2 : If $$-$$q is shifted towards right by x
$$F = {F_2} - {F_1} = {{\lambda q} \over {2\pi {\varepsilon _0}}}\left( {{1 \over {{d \over 2} - x}} - {1 \over {{d \over 2} + x}}} \right)$$ (towards right)
Thus, +q exhibits SHM while $$-$$q continues to move towards rightwards.
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