Two charges `q` and `-3q` are placed fixed on x-axis separated by distance `d`. Where should a third charge `2q` be placed such that it will not experience any force ?

Options

A. `(d-sqrt3d)/2`
B. `(d+sqrt3d)/2`
C. `(d+3d)/2`
D. `(d-3d)/2`

Explanation

To solve the problem, we need to find the position of a third charge \(2q\) such that it experiences no net force due to the other two charges \(q\) and \(-3q\) placed on the x-axis. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
We have two charges:
- Charge \(q\) is located at \(x = 0\).
- Charge \(-3q\) is located at \(x = d\).

We need to find a position \(x\) for the charge \(2q\) such that the net force acting on it is zero.

### Step 2: Determine Possible Positions
The charge \(2q\) can be placed either:
1. Between the two charges (not feasible as explained in the video).
2. To the left of charge \(q\) (at \(x < 0\)).
3. To the right of charge \(-3q\) (at \(x > d\)).

### Step 3: Analyze the Forces
Let’s consider placing \(2q\) to the left of charge \(q\) at a distance \(x\) from \(q\). The distances from \(2q\) to the two charges will be:
- Distance to \(q\): \(x\)
- Distance to \(-3q\): \(d + x\)

### Step 4: Write the Force Equations
The forces acting on \(2q\) due to the charges are:
- Force \(F_1\) due to charge \(q\) (repulsive):
\[ F_1 = k \cdot \frac{q \cdot 2q}{x^2} = \frac{2kq^2}{x^2} \]
- Force \(F_2\) due to charge \(-3q\) (attractive):
\[ F_2 = k \cdot \frac{(-3q) \cdot 2q}{(d + x)^2} = \frac{-6kq^2}{(d + x)^2} \]

### Step 5: Set Forces Equal for Equilibrium
For the charge \(2q\) to experience no net force, the magnitudes of \(F_1\) and \(F_2\) must be equal:
\[ \frac{2kq^2}{x^2} = \frac{6kq^2}{(d + x)^2} \]

### Step 6: Simplify the Equation
Cancelling \(kq^2\) from both sides gives:
\[ \frac{2}{x^2} = \frac{6}{(d + x)^2} \]
Cross-multiplying yields:
\[ 2(d + x)^2 = 6x^2 \]
Expanding and simplifying:
\[ 2(d^2 + 2dx + x^2) = 6x^2 \]
\[ 2d^2 + 4dx + 2x^2 = 6x^2 \]
\[ 2d^2 + 4dx - 4x^2 = 0 \]
Dividing through by 2:
\[ d^2 + 2dx - 2x^2 = 0 \]

### Step 7: Solve the Quadratic Equation
Rearranging gives:
\[ 2x^2 - 2dx - d^2 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 2\), \(b = -2d\), and \(c = -d^2\):
\[ x = \frac{2d \pm \sqrt{(-2d)^2 - 4 \cdot 2 \cdot (-d^2)}}{2 \cdot 2} \]
\[ x = \frac{2d \pm \sqrt{4d^2 + 8d^2}}{4} \]
\[ x = \frac{2d \pm \sqrt{12d^2}}{4} \]
\[ x = \frac{2d \pm 2\sqrt{3}d}{4} \]
\[ x = \frac{d(1 \pm \sqrt{3})}{2} \]

### Step 8: Determine Valid Position
Since \(x\) must be negative (to the left of \(q\)), we take:
\[ x = \frac{d(1 - \sqrt{3})}{2} \]

### Final Answer
The charge \(2q\) should be placed at:
\[ x = \frac{d(1 - \sqrt{3})}{2} \]