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)`

Explanation

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

### Step 1: Understanding the Configuration
We have two charges, \( +4q \) and \( +q \), placed at a distance \( L \) apart. We need to find the position and magnitude of a third charge \( Q \) such that the net force acting on it is zero.

### Step 2: Setting Up the Problem
Let's place \( +4q \) at point \( A \) and \( +q \) at point \( B \). The distance between \( A \) and \( B \) is \( L \). We will denote the position of charge \( Q \) as \( x \) from charge \( +4q \) (at point \( A \)). Therefore, the distance from charge \( Q \) to charge \( +q \) (at point \( B \)) will be \( L - x \).

### Step 3: Applying the Condition for Equilibrium
For charge \( Q \) to be in equilibrium, the net force acting on it must be zero. This means the force exerted on \( Q \) by \( +4q \) must be equal in magnitude and opposite in direction to the force exerted on \( Q \) by \( +q \).

The electrostatic force between two charges is given by Coulomb's law:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.

### Step 4: Writing the Forces
The force \( F_1 \) on charge \( Q \) due to charge \( +4q \) is:
\[ F_1 = k \frac{4q |Q|}{x^2} \]
The force \( F_2 \) on charge \( Q \) due to charge \( +q \) is:
\[ F_2 = k \frac{q |Q|}{(L - x)^2} \]

### Step 5: Setting the Forces Equal
For equilibrium, we set \( F_1 = F_2 \):
\[ k \frac{4q |Q|}{x^2} = k \frac{q |Q|}{(L - x)^2} \]

### Step 6: Simplifying the Equation
We can cancel \( k \) and \( |Q| \) (assuming \( Q \neq 0 \)):
\[ \frac{4q}{x^2} = \frac{q}{(L - x)^2} \]
Dividing both sides by \( q \) gives:
\[ \frac{4}{x^2} = \frac{1}{(L - x)^2} \]

### Step 7: Cross-Multiplying
Cross-multiplying gives:
\[ 4(L - x)^2 = x^2 \]

### Step 8: Expanding and Rearranging
Expanding the left side:
\[ 4(L^2 - 2Lx + x^2) = x^2 \]
This simplifies to:
\[ 4L^2 - 8Lx + 4x^2 = x^2 \]
Rearranging gives:
\[ 3x^2 - 8Lx + 4L^2 = 0 \]

### Step 9: Solving the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = -8L \), and \( c = 4L^2 \):
\[ x = \frac{8L \pm \sqrt{(-8L)^2 - 4 \cdot 3 \cdot 4L^2}}{2 \cdot 3} \]
\[ x = \frac{8L \pm \sqrt{64L^2 - 48L^2}}{6} \]
\[ x = \frac{8L \pm \sqrt{16L^2}}{6} \]
\[ x = \frac{8L \pm 4L}{6} \]

### Step 10: Finding the Possible Values of \( x \)
This gives us two possible solutions:
1. \( x = \frac{12L}{6} = 2L \) (not valid since it's outside the range)
2. \( x = \frac{4L}{6} = \frac{2L}{3} \)

### Step 11: Finding the Magnitude of Charge \( Q \)
Now we need to find the magnitude of \( Q \). We can substitute \( x = \frac{2L}{3} \) back into the force balance equation:
\[ F_1 = k \frac{4q |Q|}{(\frac{2L}{3})^2} = k \frac{4q |Q|}{\frac{4L^2}{9}} = k \frac{9q |Q|}{L^2} \]
\[ F_2 = k \frac{q |Q|}{(L - \frac{2L}{3})^2} = k \frac{q |Q|}{(\frac{L}{3})^2} = k \frac{9q |Q|}{L^2} \]
Setting \( F_1 = F_2 \) gives:
\[ k \frac{9q |Q|}{L^2} = k \frac{9q |Q|}{L^2} \]
This confirms that the forces are balanced.

### Conclusion
The location of the third charge \( Q \) is at \( \frac{2L}{3} \) from the charge \( +4q \), and the magnitude of the charge \( Q \) must be \( -\frac{4q}{9} \) for the system to be in equilibrium.