Two points charges `+4q` and `+q` are separated by distance r, where should a third point charge Q be placed that the whole system remains in equilibrium

Options

A. `(2r)/(3)` from q
B. `(r )/( 3)` from q
C. `(r )/( 2) ` from q
D. `(r )/( 3)` from 4q

Answer

To find the position of the third charge \( Q \) such that the entire system remains in equilibrium, we can follow these steps:

### Step 1: Define the System
We have two point charges:
- Charge \( +4q \) located at point A (let's say at position 0).
- Charge \( +q \) located at point B (let's say at position \( r \)).

We want to place a third charge \( Q \) at a distance \( x \) from charge \( +4q \) and \( r - x \) from charge \( +q \).

### Step 2: Set Up the Forces
For the system to be in equilibrium, the net force acting on each charge must be zero.

1. **Force on charge \( +4q \)**:
- The force exerted by charge \( Q \) on \( +4q \) is repulsive (since both are positive).
- The force exerted by charge \( +q \) on \( +4q \) is also repulsive.

The forces can be expressed as:
\[ F_{4q} = k \frac{4qQ}{x^2} - k \frac{q \cdot 4q}{r^2} \]

2. **Force on charge \( +q \)**:
- The force exerted by charge \( Q \) on \( +q \) is repulsive.
- The force exerted by charge \( +4q \) on \( +q \) is also repulsive.

The forces can be expressed as:
\[ F_{q} = k \frac{qQ}{(r - x)^2} - k \frac{4q \cdot q}{x^2} \]

### Step 3: Equate Forces for Equilibrium
For charge \( +4q \):
\[ k \frac{4qQ}{x^2} = k \frac{4q^2}{r^2} \]
This simplifies to:
\[ Q = \frac{q r^2}{x^2} \]

For charge \( +q \):
\[ k \frac{qQ}{(r - x)^2} = k \frac{4q^2}{x^2} \]
This simplifies to:
\[ Q = \frac{4q (r - x)^2}{x^2} \]

### Step 4: Set the Equations Equal
Now we can set the two expressions for \( Q \) equal to each other:
\[ \frac{q r^2}{x^2} = \frac{4q (r - x)^2}{x^2} \]
Cancelling \( q \) and \( x^2 \) (assuming \( q \neq 0 \) and \( x \neq 0 \)):
\[ r^2 = 4(r - x)^2 \]

### Step 5: Solve for \( x \)
Expanding and rearranging:
\[ r^2 = 4(r^2 - 2rx + x^2) \]
\[ r^2 = 4r^2 - 8rx + 4x^2 \]
\[ 0 = 3r^2 - 8rx + 4x^2 \]
This is a quadratic equation in \( x \):
\[ 4x^2 - 8rx + 3r^2 = 0 \]

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

### Step 7: Find the Values of \( x \)
This gives us two possible solutions:
1. \( x = \frac{12r}{8} = \frac{3r}{2} \) (not valid since it exceeds \( r \))
2. \( x = \frac{4r}{8} = \frac{r}{2} \)

### Step 8: Determine the Position of Charge \( Q \)
Thus, the charge \( Q \) should be placed at a distance \( x = \frac{r}{2} \) from charge \( +4q \) to maintain equilibrium.