If Two charges `+q and +4q` are separated by a distance d and a point charge Q is placed on the line joining the above two charges and in between them such that all charges are in equilibrium. Then the charge Q and it's position are

Options

A. A) `-(4q)/(9) "at a distance"(d)/(3)"from4q"`
B. B) `-(2Q)/(3) "at a distance" (d)/(3) "from q"`
C. C) `-(4q)/(9) "at a distance" (d)/(3) "from q"`
D. D) `(-2Q)/(3)"at a distance" (d)/(3) "from 4q"`

Answer

To solve the problem step by step, we need to find the charge \( Q \) and its position between the two charges \( +q \) and \( +4q \) such that the system is in equilibrium.

### Step 1: Understand the setup
We have two charges, \( +q \) and \( +4q \), separated by a distance \( d \). We need to place a charge \( Q \) between them at a distance \( x \) from the charge \( +q \).

### Step 2: Apply the equilibrium condition
For the charge \( Q \) to be in equilibrium, the net force acting on it due to the other two charges must be zero. The force exerted on \( Q \) by \( +q \) must equal the force exerted on \( Q \) by \( +4q \).

The force \( F_1 \) on \( Q \) due to \( +q \) is given by:
\[ F_1 = k \frac{q \cdot Q}{x^2} \]
where \( k \) is Coulomb's constant.

The force \( F_2 \) on \( Q \) due to \( +4q \) is given by:
\[ F_2 = k \frac{4q \cdot Q}{(d - x)^2} \]

### Step 3: Set the forces equal
For equilibrium, we set \( F_1 = F_2 \):
\[ k \frac{q \cdot Q}{x^2} = k \frac{4q \cdot Q}{(d - x)^2} \]

### Step 4: Cancel out common terms
We can cancel \( k \) and \( Q \) (assuming \( Q \neq 0 \)):
\[ \frac{q}{x^2} = \frac{4q}{(d - x)^2} \]

### Step 5: Simplify the equation
Now, we can simplify this equation:
\[ \frac{1}{x^2} = \frac{4}{(d - x)^2} \]
Cross-multiplying gives:
\[ (d - x)^2 = 4x^2 \]

### Step 6: Expand and rearrange
Expanding the left side:
\[ d^2 - 2dx + x^2 = 4x^2 \]
Rearranging gives:
\[ d^2 - 2dx - 3x^2 = 0 \]

### Step 7: Solve the quadratic equation
This is a quadratic equation in \( x \):
\[ 3x^2 + 2dx - d^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = 2d \), and \( c = -d^2 \):
\[ x = \frac{-2d \pm \sqrt{(2d)^2 - 4 \cdot 3 \cdot (-d^2)}}{2 \cdot 3} \]
\[ x = \frac{-2d \pm \sqrt{4d^2 + 12d^2}}{6} \]
\[ x = \frac{-2d \pm \sqrt{16d^2}}{6} \]
\[ x = \frac{-2d \pm 4d}{6} \]

### Step 8: Calculate the possible values of \( x \)
This gives us two possible solutions:
1. \( x = \frac{2d}{6} = \frac{d}{3} \) (valid, as it is between the charges)
2. \( x = \frac{-6d}{6} = -d \) (not valid, as it is outside the range)

Thus, the position of charge \( Q \) is \( x = \frac{d}{3} \).

### Step 9: Find the magnitude of charge \( Q \)
Now, we substitute \( x = \frac{d}{3} \) back into the force equation to find \( Q \):
Using the equilibrium condition again:
\[ k \frac{q \cdot Q}{(\frac{d}{3})^2} = k \frac{4q \cdot Q}{(d - \frac{d}{3})^2} \]
This simplifies to:
\[ \frac{q}{\frac{d^2}{9}} = \frac{4q}{(\frac{2d}{3})^2} \]
\[ \frac{9q}{d^2} = \frac{4q}{\frac{4d^2}{9}} \]
Cross-multiplying gives:
\[ 9q \cdot \frac{4d^2}{9} = 4q \cdot d^2 \]
This simplifies to:
\[ 36q = 36q \]
Thus, we find that:
\[ Q = -\frac{4q}{9} \]

### Conclusion
The charge \( Q \) is \( -\frac{4q}{9} \) and it is positioned at a distance \( \frac{d}{3} \) from the charge \( +q \).