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 solve the problem of where to place a third point charge \( Q \) such that the whole system remains in equilibrium, we can follow these steps:

### Step 1: Understand the System
We have two point charges:
- Charge 1: \( +4q \)
- Charge 2: \( +q \)

These two charges are separated by a distance \( r \). We need to find the position of a third charge \( Q \) such that the net force acting on it due to the other two charges is zero.

### Step 2: Define the Position of Charge \( Q \)
Let's denote the position of charge \( Q \) as being at a distance \( x \) from the charge \( +4q \). Consequently, the distance from charge \( +q \) will be \( r - x \).

### Step 3: Apply Coulomb's Law
According to Coulomb's law, the force between two point charges is given by:
\[ F = k \frac{|q_1 q_2|}{d^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( d \) is the distance between them.

### Step 4: Calculate the Forces Acting on Charge \( Q \)
1. The force \( F_1 \) exerted on \( Q \) by \( +4q \):
\[ F_1 = k \frac{4qQ}{x^2} \]
2. The force \( F_2 \) exerted on \( Q \) by \( +q \):
\[ F_2 = k \frac{qQ}{(r - x)^2} \]

### Step 5: Set the Forces Equal for Equilibrium
For the system to be in equilibrium, the forces \( F_1 \) and \( F_2 \) must be equal in magnitude:
\[ F_1 = F_2 \]
This gives us the equation:
\[ k \frac{4qQ}{x^2} = k \frac{qQ}{(r - x)^2} \]

### Step 6: Simplify the Equation
We can cancel \( k \) and \( qQ \) from both sides (assuming \( Q \neq 0 \) and \( q \neq 0 \)):
\[ \frac{4}{x^2} = \frac{1}{(r - x)^2} \]

### Step 7: Cross-Multiply and Rearrange
Cross-multiplying gives:
\[ 4(r - x)^2 = x^2 \]
Expanding the left side:
\[ 4(r^2 - 2rx + x^2) = x^2 \]
This simplifies to:
\[ 4r^2 - 8rx + 4x^2 = x^2 \]
Rearranging gives:
\[ 3x^2 - 8rx + 4r^2 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = -8r \), and \( c = 4r^2 \):
\[ x = \frac{8r \pm \sqrt{(-8r)^2 - 4 \cdot 3 \cdot 4r^2}}{2 \cdot 3} \]
Calculating the discriminant:
\[ 64r^2 - 48r^2 = 16r^2 \]
Thus,
\[ x = \frac{8r \pm 4r}{6} \]
This gives two possible solutions:
1. \( x = \frac{12r}{6} = 2r \) (not valid as it exceeds the distance)
2. \( x = \frac{4r}{6} = \frac{2r}{3} \)

### Step 9: Find the Distance from Charge \( +q \)
The distance from charge \( +q \) will then be:
\[ r - x = r - \frac{2r}{3} = \frac{r}{3} \]

### Conclusion
The third charge \( Q \) should be placed at a distance of \( \frac{r}{3} \) from the charge \( +q \) (or \( \frac{2r}{3} \) from the charge \( +4q \)) for the system to remain in equilibrium.