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Standard VIII

Physics

Earthing

Two charges, ...

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 $q_{0} = \frac{q}{2}$ is placed at 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

- y

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- \frac{1}{y}

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y

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\frac{1}{y}

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Solution

The correct option is C $y$

$\Rightarrow$

Net force: $\Rightarrow F_{n e t} = 2 F cos θ$

$F_{n e t} = \frac{2 k q (\frac{q}{2})}{{(\sqrt{y^{2} + a^{2}})}^{2}} . \frac{y}{\sqrt{y^{2} + a^{2}}}$

$F_{n e t} = \frac{2 k q (\frac{q}{2}) y}{{(y^{2} + a^{2})}^{\frac{3}{2}}} F_{n e t} = \frac{k q^{2} y}{a^{3}} \Rightarrow F \propto y$

Hence, $A$ is correct.

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Similar questions

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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

The correct option is C y ⇒ Net force: ⇒Fnet=2Fcosθ Fnet=2kq(q2)(√y2+a2)2.y√y2+a2 Fnet=2kq(q2)y(y2+a2)32Fnet=kq2ya3⇒F∝y Hence, A is correct.

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 $q_{0} = \frac{q}{2}$ is placed at 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