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Step-by-Step Solution
Step 1: Identify the Configuration
Two charges, each of magnitude $q$, are placed on the x-axis at $x = -a$ and $x = a$. A particle of charge $q_0 = \tfrac{q}{2}$ and mass $m$ is placed at the origin $(0,0)$. The particle is displaced slightly along the y-axis by a small distance $y$, with $y \ll a$.
Step 2: Express the Distance of $q_0$ from Each Charge
After the small displacement, the new position of $q_0$ is $(0,y)$. The distance of this point from each charge at $(\pm a, 0)$ is r = a2+y2. Since $y \ll a,$ we can use the approximate expansion for small $y$: r≈a (higher-order terms in y2 are negligible).
Step 3: Determine the Direction of Forces
The force on $q_0$ due to a charge $q$ located at $(a,0)$ has components in both the x and y directions. Similarly, the force due to the charge at $(-a,0)$ also has x and y components.
The horizontal (x-) components from the two charges are equal in magnitude and opposite in direction, so they cancel each other out.
The vertical (y-) components from the two charges add up because of symmetry.
Step 4: Calculate the Net Force in the y-Direction
Let $F_{\text{net},y}$ be the net force in the y-direction. The magnitude of the force by one charge on $q_0$ is given by Coulomb’s law: F=kq q0r2, where $k$ is the Coulomb constant. For a small displacement along $y$, the y-component of each force is proportional to $y$ (since $\sin\theta \approx \tfrac{y}{r}$ for small $y$). Because $r \approx a$, each force’s y-component behaves like $\propto y/a^2,$ and adding both contributions gives a total proportional to $y.$
Step 5: Conclude the Dependence on $y$
Since the net force in the y-direction on the particle is directly proportional to $y$ for small $y \ll a,$ the correct answer is: Net force ∝ y.
Final Answer
The net force acting on the particle is proportional to $y$.
