Two charges- each equal to q- are kept x-a and x-a on the x-axis- A particle of mass m and charge q-0-frac-q-2- is placed the origin- If charge q-0- is given a small displacement -y - a- along the y-axis- the net force acting on the particle is proportional to-frac-1-y-2-y-frac-1-y-y

Question

Toppr Admin · Accepted Answer

Fnet-2Fcos-x3B8-2k-x22C5-q-x22C5-q-2-x221A-a2-y2-2-x22C5-y-x221A-a2-y2-kq2y-a2-y2-3-2

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=q2 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:1y2−y−1yy

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

$F_{n e t} = 2 F cos θ$ $= 2 \frac{k \cdot q \cdot q / 2}{(\sqrt{a^{2} + y^{2}})^{2}} \cdot \frac{y}{\sqrt{a^{2} + y^{2}}}$ $= \frac{k q^{2} y}{{(a^{2} + y^{2})}^{3 / 2}}$