Two electric `+q and +4q` are placed at a distance 6a apart on a horizontal plane. Find the position of the point on the line joining the two charges where the electric field is zero.

Explanation

To find the position on the line joining the two charges \(+q\) and \(+4q\) where the electric field is zero, follow these steps:

1. **Identify the positions and distances:**
- Let the charge \(+q\) be at position \(A\).
- Let the charge \(+4q\) be at position \(B\).
- The distance between \(A\) and \(B\) is \(6a\).

2. **Determine the potential zero electric field point:**
- The electric field due to a charge is given by \(E = \frac{kQ}{r^2}\), where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge.
- The electric field will be zero at a point between the charges where the fields due to each charge cancel each other out.

3. **Assume the point \(P\) is at a distance \(x\) from \(+q\):**
- The distance from \(+4q\) to \(P\) will be \(6a - x\).

4. **Set up the equation for the electric fields:**
- The electric field due to \(+q\) at point \(P\) is \(E_1 = \frac{kq}{x^2}\).
- The electric field due to \(+4q\) at point \(P\) is \(E_2 = \frac{k(4q)}{(6a - x)^2}\).

5. **Set the magnitudes of the electric fields equal to each other:**
\[ \frac{kq}{x^2} = \frac{k(4q)}{(6a - x)^2} \]

6. **Simplify the equation:**
\[ \frac{q}{x^2} = \frac{4q}{(6a - x)^2} \]
\[ \frac{1}{x^2} = \frac{4}{(6a - x)^2} \]

7. **Cross-multiply to solve for \(x\):**
\[ (6a - x)^2 = 4x^2 \]

8. **Expand and simplify:**
\[ 36a^2 - 12ax + x^2 = 4x^2 \]
\[ 36a^2 - 12ax + x^2 = 4x^2 \]
\[ 36a^2 - 12ax = 3x^2 \]

9. **Rearrange the equation:**
\[ 3x^2 + 12ax - 36a^2 = 0 \]

10. **Solve the quadratic equation:**
\[ x^2 + 4ax - 12a^2 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4a\), and \(c = -12a^2\):
\[ x = \frac{-4a \pm \sqrt{(4a)^2 - 4 \cdot 1 \cdot (-12a^2)}}{2 \cdot 1} \]
\[ x = \frac{-4a \pm \sqrt{16a^2 + 48a^2}}{2} \]
\[ x = \frac{-4a \pm \sqrt{64a^2}}{2} \]
\[ x = \frac{-4a \pm 8a}{2} \]

11. **Calculate the possible values for \(x\):**
\[ x = \frac{-4a + 8a}{2} = \frac{4a}{2} = 2a \]
\[ x = \frac{-4a - 8a}{2} = \frac{-12a}{2} = -6a \]

12. **Select the positive value:**
- Since \(x = -6a\) is not within the range between the charges, we discard it.
- Therefore, \(x = 2a\).

**Conclusion:**
The electric field is zero at a point \(2a\) from the charge \(+q\) on the line joining the two charges.