A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.50 m. A third charge of +2 q is located on the x axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located ?

Options

A. `+0.50 m`
B. `+0.71 m`
C. `-0.50 m`
D. `-0.71 m`

Answer

To solve the problem, we need to determine the position of a third charge \( +2q \) on the x-axis such that the net electrostatic force on the charge at the origin (which is \( +q \)) doubles, while keeping the direction of the force unchanged.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Charge at the Origin:**
- The charge at the origin \( +q \) experiences a repulsive force due to the charge at \( x = 0.5 \, m \) (also \( +q \)).
- The force \( F_1 \) between the two charges \( +q \) at the origin and \( +q \) at \( x = 0.5 \, m \) is given by Coulomb's law:
\[ F_1 = k \frac{q \cdot q}{(0.5)^2} = k \frac{q^2}{0.25} = 4kq^2 \]
- This force acts to the left (negative x-direction).

2. **Introduce the Third Charge:**
- Let the position of the third charge \( +2q \) be at a distance \( r \) from the origin on the x-axis. Thus, its position is \( x = r \).
- The force \( F_2 \) on the charge at the origin due to the charge \( +2q \) is:
\[ F_2 = k \frac{q \cdot 2q}{r^2} = k \frac{2q^2}{r^2} \]
- This force acts to the right (positive x-direction).

3. **Set Up the Equation for the Net Force:**
- The net force \( F_{net} \) on the charge at the origin is the sum of the forces \( F_1 \) and \( F_2 \):
\[ F_{net} = F_2 - F_1 = k \frac{2q^2}{r^2} - 4kq^2 \]
- We want this net force to double the initial force \( F_1 \):
\[ F_{net} = 2F_1 = 2(4kq^2) = 8kq^2 \]
- Therefore, we set up the equation:
\[ k \frac{2q^2}{r^2} - 4kq^2 = 8kq^2 \]

4. **Simplify the Equation:**
- Dividing through by \( kq^2 \) (assuming \( k \) and \( q \) are not zero):
\[ \frac{2}{r^2} - 4 = 8 \]
- Rearranging gives:
\[ \frac{2}{r^2} = 12 \]
- Thus:
\[ r^2 = \frac{2}{12} = \frac{1}{6} \]
- Taking the square root:
\[ r = \frac{1}{\sqrt{6}} \approx 0.408 \, m \]

5. **Calculate the Total Distance from the Origin:**
- The total distance from the origin to the third charge is:
\[ 0.5 \, m + r \approx 0.5 + 0.408 \approx 0.908 \, m \]

### Final Answer:
The third charge \( +2q \) should be located at approximately \( 0.908 \, m \) from the origin on the x-axis.