Two charge q and -3q are placed fixed on X-axis at a separation ‘d’. Where should a third charge 2q be placed such that it will not experience any force. Also calculate potential energy of the system.

To solve the problem of where to place a third charge \(2q\) so that it experiences no net force due to the other two charges \(q\) and \(-3q\), we need to analyze the forces acting on the charge \(2q\) based on Coulomb's law. Let's break this down step by step.

Understanding the Charge Configuration

We have two fixed charges on the X-axis: charge \(q\) is located at the origin (0,0), and charge \(-3q\) is positioned at (d,0). The goal is to find a point along the X-axis where the third charge \(2q\) can be placed such that the forces from \(q\) and \(-3q\) cancel each other out.

Identifying Possible Locations

The third charge \(2q\) can be placed in three potential regions:

• To the left of charge \(q\) (at \(x < 0\))
• Between charges \(q\) and \(-3q\) (at \(0 < x < d\))
• To the right of charge \(-3q\) (at \(x > d\))

Analyzing Forces

We will analyze the forces in each region to find where the net force on \(2q\) is zero.

1. To the Left of Charge \(q\) (x < 0)

In this region, both charges \(q\) and \(-3q\) will exert forces on \(2q\) that are directed to the right. Hence, there will be no point where the forces cancel out.

2. Between Charges \(q\) and \(-3q\) (0 < x < d)

Here, the force due to charge \(q\) will be directed to the right (repulsive), while the force due to charge \(-3q\) will be directed to the left (attractive). The forces will not balance out because the magnitude of the force from \(-3q\) will always be greater than that from \(q\) in this region.

3. To the Right of Charge \(-3q\) (x > d)

In this region, both charges will exert forces on \(2q\) that are directed to the left. However, we can find a point where these forces balance out. Let's denote the distance from charge \(-3q\) to the charge \(2q\) as \(x - d\).

Setting Up the Force Balance Equation

Using Coulomb's law, the force \(F_1\) exerted by charge \(q\) on \(2q\) is:

F₁ = k \frac{q \cdot 2q}{x^2}

And the force \(F_2\) exerted by charge \(-3q\) on \(2q\) is:

F₂ = k \frac{(-3q) \cdot 2q}{(x - d)^2}

For the net force to be zero:

F₁ + F₂ = 0

Substituting the expressions for \(F_1\) and \(F_2\):

k \frac{q \cdot 2q}{x^2} = k \frac{3q \cdot 2q}{(x - d)^2}

We can simplify this equation by canceling \(k\) and \(2q\) (assuming \(q \neq 0\)):

\frac{q}{x^2} = \frac{3q}{(x - d)^2}

Canceling \(q\) gives:

\frac{1}{x^2} = \frac{3}{(x - d)^2}

Cross-multiplying leads to:

(x - d)^2 = 3x^2

Expanding and rearranging:

x^2 - 2dx + d^2 = 3x^2

2x^2 - 2dx + d^2 = 0

Dividing through by 2 gives:

x^2 - dx + \frac{d^2}{2} = 0

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

x = \frac{d \pm \sqrt{d^2 - 2d^2}}{2} = \frac{d \pm \sqrt{-d^2}}{2}

This indicates that the charge \(2q\) should be placed at:

x = \frac{d}{2} + \frac{d\sqrt{2}}{2} \text{ or } x = \frac{d}{2} - \frac{d\sqrt{2}}{2}

Calculating Potential Energy of the System

The potential energy \(U\) of a system of point charges can be calculated using the formula:

U = k \sum_{i < j} \frac{q_i q_j}{r_{ij}}

For our three charges, the potential energy contributions are:

• Between \(q\) and \(2q\): \(U_{q,2q} = k \frac{q \cdot 2q}{r_{q,2q}} = k \frac{2q^2}{x}\)
• Between \(-3q\) and \(2q\): \(U_{-3q,2q} = k \frac{-3q \cdot 2q}{r_{-3q,2q}} = -k \frac{6q^2}{x - d}\)
• Between \(q\) and \(-3q\): \(U_{q,-3q} = k \frac{q \cdot (-3q)}{d} = -k