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 $y ( y << a)$ along the y-axis, the net force acting on the particle is proportional to

• A. $y$
• B. $-y$
• C. $\frac{1}{y}$
• D. $-\frac{1}{y}$

Answer 1

The Correct Option is A

To solve this problem, we need to find the net force acting on the charge \(q_0\) when it is given a small displacement \(y\) along the y-axis.

Consider two point charges, each of magnitude \(q\), placed at \(x = -a\) and \(x = a\) on the x-axis. These charges create an electric field in the space around them. A small test charge \(q_0 = \frac{q}{2}\) (positive) is placed at the origin and displaced slightly along the y-axis.

The electric field due to a point charge \(q\) at a distance \(r\) is given by:

\(E = \frac{k \cdot q}{r^2}\)

For a point charge of mass \(m\) and charge \(q_0\) displaced on the y-axis, we have:

The distance from charge \(q_0\) to each charge \(q\) is approximately \(d = \sqrt{a^2 + y^2}\).

As \(y \ll a\), we can use the binomial expansion to approximate:

\(d \approx a \left( 1 + \frac{y^2}{2a^2} \right)\)

The electric force \(F\) on \(q_0\) due to each charge \(q\) is:

\(F = \frac{k \cdot q \cdot q_0}{d^2}\)

Since \(q_0\) is only slightly displaced along the y-axis, the net force will primarily have a y-component. The y-component of the force due to each charge is approximately:

\(F_y \approx \frac{k \cdot q \cdot q_0 \cdot y}{a^3}\)

The forces due to both charges are directed oppositely along the y-axis, and thus cancel each other out horizontally but contribute additively along the y-axis. Therefore, the net force \(F_{\text{net}, y}\) is twice the y-component of this force due to symmetry:

\(F_{\text{net}, y} = 2 \cdot F_y\)

\(F_{\text{net}, y} \approx 2 \cdot \frac{k \cdot q \cdot \left(\frac{q}{2}\right) \cdot y}{a^3} = \frac{k \cdot q^2 \cdot y}{a^3}\)

Hence, the net force on \(q_0\) is proportional to \(y\), as the horizontal components cancel out due to symmetry.

Therefore, the correct answer is \(y\).