Two point charges `+4q` and `+q` are placed at a distance `L` apart. `A` third charge `Q` is so placed that all the three charges are in equilibrium. Then location. And magnitude of the third charge will be

Options

A. At a distance `(L)/(3)` from `+4q` charge, `(4q)/(9)`
B. At a distance `(L)/(3)` from +4q charge, `-(4q)/(9)`
C. At a distance `(2L)/(3)` from +4q charge, `-(4q)/(9)`
D. At a distance `(2L)/(3)` from +q charge, `(4q)/(9)`

Answer

To solve the problem of finding the location and magnitude of the third charge \( Q \) such that the three charges \( +4q \), \( +q \), and \( Q \) are in equilibrium, we can follow these steps:

### Step 1: Understand the Configuration
We have two charges:
- Charge \( +4q \) located at point A.
- Charge \( +q \) located at point B, a distance \( L \) apart from \( +4q \).

We need to place a third charge \( Q \) between these two charges such that all three charges are in equilibrium.

### Step 2: Identify the Position of Charge \( Q \)
Let’s denote the distance from charge \( +q \) to charge \( Q \) as \( x \). Therefore, the distance from charge \( +4q \) to charge \( Q \) will be \( L - x \).

### Step 3: Set Up the Force Equilibrium Condition
For charge \( Q \) to be in equilibrium, the net force acting on it must be zero. This means that the force exerted on \( Q \) by \( +q \) must equal the force exerted on \( Q \) by \( +4q \).

The forces can be expressed using Coulomb's law:
- The force \( F_1 \) due to charge \( +q \) on charge \( Q \):
\[ F_1 = k \frac{q \cdot Q}{x^2} \]
- The force \( F_2 \) due to charge \( +4q \) on charge \( Q \):
\[ F_2 = k \frac{4q \cdot Q}{(L - x)^2} \]

Setting these forces equal for equilibrium:
\[ k \frac{q \cdot Q}{x^2} = k \frac{4q \cdot Q}{(L - x)^2} \]

### Step 4: Simplify the Equation
We can cancel \( k \) and \( Q \) (assuming \( Q \neq 0 \)):
\[ \frac{q}{x^2} = \frac{4q}{(L - x)^2} \]

This simplifies to:
\[ \frac{1}{x^2} = \frac{4}{(L - x)^2} \]

### Step 5: Cross Multiply and Solve for \( x \)
Cross multiplying gives:
\[ (L - x)^2 = 4x^2 \]

Expanding the left side:
\[ L^2 - 2Lx + x^2 = 4x^2 \]

Rearranging the equation:
\[ L^2 - 2Lx - 3x^2 = 0 \]

### Step 6: Use the Quadratic Formula
This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where:
- \( a = -3 \)
- \( b = -2L \)
- \( c = L^2 \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{2L \pm \sqrt{(-2L)^2 - 4(-3)(L^2)}}{2(-3)} \]
\[ x = \frac{2L \pm \sqrt{4L^2 + 12L^2}}{-6} \]
\[ x = \frac{2L \pm \sqrt{16L^2}}{-6} \]
\[ x = \frac{2L \pm 4L}{-6} \]

Calculating the two possible values:
1. \( x = \frac{6L}{-6} = -L \) (not valid as distance cannot be negative)
2. \( x = \frac{-2L}{-6} = \frac{L}{3} \)

### Step 7: Determine the Magnitude of Charge \( Q \)
Now, we need to find the charge \( Q \). We know that:
\[ F_{4q} = F_{Q} \]
Using the force equations again:
\[ k \frac{4q \cdot Q}{(L - \frac{L}{3})^2} = k \frac{q \cdot Q}{(\frac{L}{3})^2} \]
This simplifies to:
\[ \frac{4q}{(\frac{2L}{3})^2} = \frac{q}{(\frac{L}{3})^2} \]
Solving this gives:
\[ 4 \cdot \frac{9}{4} = 1 \]
Thus, \( Q = -\frac{4q}{9} \).

### Final Result
The third charge \( Q \) should be placed at a distance \( \frac{L}{3} \) from charge \( +q \) (or \( \frac{2L}{3} \) from charge \( +4q \)), and its magnitude should be \( -\frac{4q}{9} \).