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. (a) `x/2`
B. (b) `x/3`
C. (c) `x/6`
D. (d) `(2x)/3`

Answer

To solve the problem, we need to find the distance \( r \) from the charge \( +e \) at which the charge \( q \) must be placed so that it is in equilibrium. The two charges \( +4e \) and \( +e \) are separated by a distance \( x \).

### Step-by-Step Solution:

1. **Identify the Charges and Distances**:
- Let \( Q_1 = +4e \) (located at one end).
- Let \( Q_2 = +e \) (located at the other end).
- The distance between \( Q_1 \) and \( Q_2 \) is \( x \).
- Let \( r \) be the distance from \( Q_2 \) (charge \( +e \)) to the charge \( q \).
- Therefore, the distance from \( Q_1 \) to \( q \) will be \( x - r \).

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

3. **Equilibrium Condition**:
- For charge \( q \) to be in equilibrium, the net force acting on it must be zero. Thus, the forces must be equal:
\[ F_{q,4e} = F_{q,e} \]
- This gives us the equation:
\[ k \frac{(4e)q}{(x - r)^2} = k \frac{eq}{r^2} \]

4. **Cancel Common Terms**:
- We can cancel \( k \) and \( q \) from both sides (assuming \( q \neq 0 \)):
\[ \frac{4e}{(x - r)^2} = \frac{e}{r^2} \]

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

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

7. **Expand and Rearrange**:
- Expanding the right side:
\[ 4r^2 = x^2 - 2xr + r^2 \]
- Rearranging the equation:
\[ 4r^2 - r^2 + 2xr - x^2 = 0 \]
\[ 3r^2 + 2xr - x^2 = 0 \]

8. **Solve the Quadratic Equation**:
- Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = 2x \), and \( c = -x^2 \).
\[ r = \frac{-2x \pm \sqrt{(2x)^2 - 4 \cdot 3 \cdot (-x^2)}}{2 \cdot 3} \]
\[ r = \frac{-2x \pm \sqrt{4x^2 + 12x^2}}{6} \]
\[ r = \frac{-2x \pm \sqrt{16x^2}}{6} \]
\[ r = \frac{-2x \pm 4x}{6} \]

9. **Calculate the Possible Values of r**:
- This gives two possible solutions:
\[ r = \frac{2x}{6} = \frac{x}{3} \quad \text{(taking the positive root)} \]
\[ r = \frac{-6x}{6} = -x \quad \text{(not physically meaningful)} \]

10. **Conclusion**:
- The distance \( r \) from charge \( +e \) where charge \( q \) must be placed for equilibrium is:
\[ r = \frac{x}{3} \]