Two 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 is in equilibrium, follow these steps:

### Step 1: Understanding the Configuration
We have two point charges, \( +q \) and \( +4q \), separated by a distance \( l \). We need to place a third charge \( q_1 \) such that the net force acting on all charges is zero.

### Step 2: Positioning the Third Charge
Assume the charge \( q_1 \) is placed at a distance \( x \) from charge \( +q \). Consequently, the distance from charge \( +4q \) will be \( l - x \).

### Step 3: Forces Acting on Charge \( q_1 \)
- The force \( F_q \) acting on \( q_1 \) due to charge \( +q \) will be directed away from \( +q \) (to the right).
- The force \( F_{4q} \) acting on \( q_1 \) due to charge \( +4q \) will be directed towards \( +4q \) (to the left).

### Step 4: Setting Up the Equilibrium Condition
For the system to be in equilibrium, the magnitudes of the forces must be equal:
\[ F_q = F_{4q} \]

Using Coulomb's Law, we can express these forces:
\[ F_q = k \frac{q_1 \cdot q}{x^2} \]
\[ F_{4q} = k \frac{q_1 \cdot 4q}{(l - x)^2} \]

### Step 5: Equating the Forces
Setting the magnitudes equal gives:
\[ k \frac{q_1 \cdot q}{x^2} = k \frac{q_1 \cdot 4q}{(l - x)^2} \]

Canceling \( k \) and \( q_1 \) (assuming \( q_1 \neq 0 \)):
\[ \frac{q}{x^2} = \frac{4q}{(l - x)^2} \]

### Step 6: Simplifying the Equation
We can simplify this equation:
\[ \frac{1}{x^2} = \frac{4}{(l - x)^2} \]
Cross-multiplying gives:
\[ (l - x)^2 = 4x^2 \]

### Step 7: Expanding and Rearranging
Expanding the left side:
\[ l^2 - 2lx + x^2 = 4x^2 \]
Rearranging gives:
\[ l^2 - 2lx - 3x^2 = 0 \]

### Step 8: Solving the Quadratic Equation
This is a quadratic equation in \( x \):
\[ 3x^2 + 2lx - l^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-2l \pm \sqrt{(2l)^2 - 4 \cdot 3 \cdot (-l^2)}}{2 \cdot 3} \]
\[ x = \frac{-2l \pm \sqrt{4l^2 + 12l^2}}{6} \]
\[ x = \frac{-2l \pm \sqrt{16l^2}}{6} \]
\[ x = \frac{-2l \pm 4l}{6} \]

Calculating the two possible values:
1. \( x = \frac{2l}{6} = \frac{l}{3} \) (valid)
2. \( x = \frac{-6l}{6} = -l \) (not valid)

### Step 9: Finding the Position of Charge \( q_1 \)
Thus, \( q_1 \) is located at \( x = \frac{l}{3} \) from charge \( +q \), which means it is \( \frac{2l}{3} \) from charge \( +4q \).

### Step 10: Determining the Magnitude and Sign of Charge \( q_1 \)
Next, we need to find the magnitude and sign of \( q_1 \). Using the equilibrium condition on charge \( +4q \):
\[ F_{4q} = F_q \]
\[ \frac{4q}{l^2} = \frac{q_1}{(l - \frac{l}{3})^2} \]
Substituting \( l - \frac{l}{3} = \frac{2l}{3} \):
\[ \frac{4q}{l^2} = \frac{q_1}{(\frac{2l}{3})^2} \]
\[ \frac{4q}{l^2} = \frac{q_1}{\frac{4l^2}{9}} \]
Cross-multiplying gives:
\[ 4q \cdot \frac{4l^2}{9} = q_1 \cdot l^2 \]
\[ q_1 = \frac{16q}{9} \]

Since \( q_1 \) must be negative to ensure equilibrium (as it counteracts the repulsive forces of the other two positive charges), we conclude:
\[ q_1 = -\frac{16q}{9} \]

### Final Answer
The third charge \( q_1 \) should be placed at a distance \( \frac{l}{3} \) from charge \( +q \), and its magnitude is \( \frac{16q}{9} \) with a negative sign.

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