⇒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 q_0=- q 2 is placed at the origin. If…
JEE Main 2013 — Physics Waves & Oscillations
2013mcqhard
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 :
Official previous-year question
Held on 7 Apr 2013 · Verified 6 Jul 2026.
Options
- A
y1
- B
−y1
- C
-y
- D
y
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