

⇒Fnet=2Fcosθ
Fnet=(y2+a2)22kq(2q)(y2+a2y)
Fnet=(y2+a2)3/22kq(2q)y≈(a3kq2)y⇒F∝(−y)
Net force F is opposite the direction of displacement y. Hence, negative sign is used.
Two charges, each equal to q, are kept at x=−a and x=a on the x-axis. A particle of mass m and charge q0=−2q is placed at the origin. If charge q0 is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to :
Held on 7 Apr 2013 · Verified 6 Jul 2026.
y1
−y1
-y
y
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Work through every JEE Main Waves & Oscillations PYQ, year by year.