Two charges `+4e` and `+e` are at a distance `x` apart. At what distance,a charge `q` must be placed from charge `+e` so that is in equilibrium

Options

A. `x//2`
B. `2x//3`
C. `x//3`
D. `x//6`

Explanation

To solve the problem of finding the distance at which a charge \( q \) must be placed from charge \( +e \) so that it is in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- We have two charges: \( +4e \) and \( +e \).
- The distance between these two charges is \( x \).

2. **Define the Position of Charge \( q \)**:
- Let the distance of charge \( q \) from charge \( +e \) be \( r \).
- Consequently, the distance from charge \( +4e \) will be \( x - r \).

3. **Set Up the Force Equations**:
- The force on charge \( q \) due to charge \( +4e \) is given by Coulomb's law:
\[ F_{4e} = k \frac{(4e)(q)}{(x - r)^2} \]
- The force on charge \( q \) due to charge \( +e \) is:
\[ F_{e} = k \frac{(e)(q)}{r^2} \]

4. **Equilibrium Condition**:
- For charge \( q \) to be in equilibrium, the magnitudes of the forces must be equal:
\[ F_{4e} = F_{e} \]
- Therefore, we can set up the equation:
\[ k \frac{(4e)(q)}{(x - r)^2} = k \frac{(e)(q)}{r^2} \]

5. **Cancel Common Terms**:
- Since \( k \) and \( q \) are common on both sides, we can cancel them out:
\[ \frac{4e}{(x - r)^2} = \frac{e}{r^2} \]

6. **Simplify the Equation**:
- Dividing both sides by \( e \) gives:
\[ \frac{4}{(x - r)^2} = \frac{1}{r^2} \]

7. **Cross Multiply**:
- Cross multiplying yields:
\[ 4r^2 = (x - r)^2 \]

8. **Expand the Right Side**:
- Expanding gives:
\[ 4r^2 = x^2 - 2xr + r^2 \]

9. **Rearranging the Equation**:
- Bringing all terms to one side results in:
\[ 4r^2 - r^2 + 2xr - x^2 = 0 \]
- This simplifies to:
\[ 3r^2 + 2xr - x^2 = 0 \]

10. **Using the Quadratic Formula**:
- We can solve this quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = 2x \), and \( c = -x^2 \).
- Thus:
\[ r = \frac{-2x \pm \sqrt{(2x)^2 - 4 \cdot 3 \cdot (-x^2)}}{2 \cdot 3} \]
\[ r = \frac{-2x \pm \sqrt{4x^2 + 12x^2}}{6} \]
\[ r = \frac{-2x \pm \sqrt{16x^2}}{6} \]
\[ r = \frac{-2x \pm 4x}{6} \]

11. **Calculating the Values of \( r \)**:
- This gives us two possible solutions:
- \( r = \frac{2x}{6} = \frac{x}{3} \) (valid, as it is positive)
- \( r = \frac{-6x}{6} = -x \) (not valid, as distance cannot be negative)

12. **Final Answer**:
- Therefore, the distance \( r \) at which charge \( q \) must be placed from charge \( +e \) for equilibrium is:
\[ r = \frac{2x}{3} \]