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Question:

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 = q/2 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:

1/y^2

y

-y

√y

Solution:

Fnet = 2Fcosθ = 2k⋅q⋅q/2(√(a²+y²))² ⋅ y/√(a²+y²) = kq²y/(a²+y²)³/²