Two force point charges +q and +4q are a distance l apart. A third charge `q_(1)` is so placed that the entire system is in equilibrium. Find the location, magnitude and sign of the third charge.

Options

A. `q_(1)=(4)/(9)q`
B. `q_(1)=(-4)/(9)q`
C. `q_(1)` is to be placed at a distance `l//3` from q.
D. `q_(1)` is to be placed at a distance `(2l)/(3)` from 4q.

Answer

To solve the problem of finding the location, magnitude, and sign of the third charge \( q_1 \) such that the entire system of charges is in equilibrium, we can follow these steps:

### Step 1: Understand the Configuration
We have two point charges: \( +q \) and \( +4q \) placed a distance \( l \) apart. We need to place a third charge \( q_1 \) in such a way that the net force on it due to the other two charges is zero.

### Step 2: Position the Charges
Let's place the charge \( +q \) at position \( 0 \) and the charge \( +4q \) at position \( l \) on a straight line. We will denote the position of \( q_1 \) as \( x \), where \( 0 < x < l \).

### Step 3: Calculate the Forces Acting on \( q_1 \)
The forces acting on \( q_1 \) due to the charges \( +q \) and \( +4q \) are given by Coulomb's law:

1. **Force due to \( +q \)**:
\[ F_1 = k \frac{q \cdot q_1}{x^2} \]
This force is directed away from \( +q \) (to the right).

2. **Force due to \( +4q \)**:
\[ F_2 = k \frac{4q \cdot q_1}{(l - x)^2} \]
This force is directed away from \( +4q \) (to the left).

### Step 4: Set Up the Equilibrium Condition
For the system to be in equilibrium, the net force on \( q_1 \) must be zero:
\[ F_1 = F_2 \]
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ k \frac{q \cdot q_1}{x^2} = k \frac{4q \cdot q_1}{(l - x)^2} \]
We can cancel \( k \) and \( q_1 \) (assuming \( q_1 \neq 0 \)):
\[ \frac{1}{x^2} = \frac{4}{(l - x)^2} \]

### Step 5: Cross-Multiply and Rearrange
Cross-multiplying gives:
\[ (l - x)^2 = 4x^2 \]
Expanding both sides:
\[ l^2 - 2lx + x^2 = 4x^2 \]
Rearranging terms:
\[ l^2 - 2lx - 3x^2 = 0 \]

### Step 6: Solve the Quadratic Equation
This is a standard quadratic equation in \( x \):
\[ 3x^2 + 2lx - l^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-2l \pm \sqrt{(2l)^2 - 4 \cdot 3 \cdot (-l^2)}}{2 \cdot 3} \]
\[ x = \frac{-2l \pm \sqrt{4l^2 + 12l^2}}{6} \]
\[ x = \frac{-2l \pm \sqrt{16l^2}}{6} \]
\[ x = \frac{-2l \pm 4l}{6} \]

### Step 7: Calculate the Possible Values of \( x \)
Calculating the two possible values:
1. \( x = \frac{2l}{6} = \frac{l}{3} \) (valid)
2. \( x = \frac{-6l}{6} = -l \) (not valid, as distance cannot be negative)

Thus, the position of \( q_1 \) is \( x = \frac{l}{3} \).

### Step 8: Determine the Magnitude and Sign of \( q_1 \)
To find the magnitude and sign of \( q_1 \), we can use the equilibrium condition again. Since \( F_1 + F_2 = 0 \), we can express this as:
\[ k \frac{q \cdot q_1}{(\frac{l}{3})^2} + k \frac{4q \cdot q_1}{(\frac{2l}{3})^2} = 0 \]
This simplifies to:
\[ \frac{q \cdot q_1}{\frac{l^2}{9}} + \frac{4q \cdot q_1}{\frac{4l^2}{9}} = 0 \]
\[ 9q \cdot q_1 + 9q \cdot q_1 = 0 \]
This implies:
\[ q_1 = -\frac{4q}{9} \]
Thus, the third charge \( q_1 \) must be negative.

### Final Answer
- **Location of \( q_1 \)**: \( \frac{l}{3} \) from \( +q \)
- **Magnitude of \( q_1 \)**: \( \frac{4q}{9} \)
- **Sign of \( q_1 \)**: Negative