Two point charges +9e and +e are kept at a distance 'a' from each other. A third charge is placed at a distance 'x' from +9e on the line joining the above two charges. For the third charge to be in equilibrium 'x' is

Options

A. `3a//4`
B. `3a//2`
C. `a//2`
D. `a//4`

Explanation

To solve this problem, we need to find the position \( x \) where a third charge \( -q \) should be placed such that it is in equilibrium due to the forces exerted by the charges \( +9e \) and \( +e \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Third Charge:**
- The third charge \( -q \) will experience an attractive force towards \( +9e \) and a repulsive force from \( +e \).

2. **Assume the Position of the Third Charge:**
- Let the third charge \( -q \) be placed at a distance \( x \) from the charge \( +9e \). Therefore, the distance from \( +e \) will be \( a - x \).

3. **Write the Force Equations:**
- The force due to \( +9e \) on \( -q \):
\[ F_1 = k \frac{9e \cdot q}{x^2} \]
- The force due to \( +e \) on \( -q \):
\[ F_2 = k \frac{e \cdot q}{(a - x)^2} \]

4. **Set the Forces Equal for Equilibrium:**
- For the third charge to be in equilibrium, the magnitudes of these forces must be equal:
\[ k \frac{9e \cdot q}{x^2} = k \frac{e \cdot q}{(a - x)^2} \]

5. **Simplify the Equation:**
- Cancel out the common terms \( k \) and \( q \):
\[ \frac{9e}{x^2} = \frac{e}{(a - x)^2} \]
- Simplify further:
\[ \frac{9}{x^2} = \frac{1}{(a - x)^2} \]

6. **Solve for \( x \):**
- Cross-multiply to solve for \( x \):
\[ 9(a - x)^2 = x^2 \]
- Expand and simplify:
\[ 9(a^2 - 2ax + x^2) = x^2 \]
\[ 9a^2 - 18ax + 9x^2 = x^2 \]
\[ 9a^2 - 18ax + 8x^2 = 0 \]

7. **Solve the Quadratic Equation:**
- This is a quadratic equation in \( x \):
\[ 8x^2 - 18ax + 9a^2 = 0 \]
- Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{18a \pm \sqrt{(18a)^2 - 4 \cdot 8 \cdot 9a^2}}{2 \cdot 8} \]
\[ x = \frac{18a \pm \sqrt{324a^2 - 288a^2}}{16} \]
\[ x = \frac{18a \pm \sqrt{36a^2}}{16} \]
\[ x = \frac{18a \pm 6a}{16} \]
- This gives two solutions:
\[ x = \frac{24a}{16} = \frac{3a}{2} \]
\[ x = \frac{12a}{16} = \frac{3a}{4} \]

8. **Select the Valid Solution:**
- Since \( x \) must be less than \( a \) (as it is the distance from \( +9e \)), the valid solution is:
\[ x = \frac{3a}{4} \]

### Final Answer:
The distance \( x \) from the charge \( +9e \) where the third charge \( -q \) should be placed to be in equilibrium is \( \boxed{\frac{3a}{4}} \).